Question 13 Marks
In Exercises $3.13$ and $3.14$, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
AnswerInstantaneous velocity is given by the first derivative of distance with respect to time i.e., $V_{in} = dx/dt$ Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time. Therefore, instantaneous speed is always equal to instantaneous velocity.
View full question & answer→Question 23 Marks
A jet airplane travelling at the speed of $500 km\ h^{–1}$ ejects its products of combustion at the speed of $1500km\ h^{–1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
AnswerSpeed of the jet airplane, $V_{j e t}=500 \mathrm{~km} / \mathrm{h}$ Relative speed of its products of combustion with respect to the plane, $V_{\text {smoke }}=-1500 \mathrm{~km} / \mathrm{h}$ Speed of its products of combustion with respect to the ground $=\mathrm{V}_{\text {smoke }}^{\prime}$ Relative speed of its products of combustion with respect to the airplane, $\mathrm{V}_{\text {smoke }}=\mathrm{V}_{\text {smoke }}^{\prime}-\mathrm{V}_{\text {jet }}-1500=\mathrm{V}_{\text {smoke }}^{\prime}-500 \mathrm{~V}_{\text {smoke }}^{\prime}=-$ $1000 \mathrm{~km} / \mathrm{h}$ The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.
View full question & answer→Question 33 Marks
A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$. To what height does the ball rise and after how long does the ball return to the player's hands? (Take $\mathrm{g}=9.8 \mathrm{~m}^{\mathrm{s}-2}$ and neglect air resistance).
AnswerInitial velocity of the ball, $u=29.4 \mathrm{~m} / \mathrm{s}$ Final velocity of the ball, $v=0$ (At maximum height, the velocity of the ball becomes zero) Acceleration, $a=-g=-9.8 \mathrm{~m} / \mathrm{s}^2$ From third equation of motion, height ( s ) can be calculated as: $\mathrm{v}^2-\mathrm{u}^2$ $=2 g s s=\left(v^2-u^2\right) / 2 g=\left((0)^2-(29.4)^2\right) / 2 \times(-9.8)=3 s$ Time of ascent $=$ Time of descent Hence, the total time taken by the ball to return to the player's hands $=3+3=6 \mathrm{~s}$.
View full question & answer→Question 43 Marks
The velocity-time graph of a particle in one-dimensional motion is shown in:
Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$:
a. $x\left(t_2\right)=x\left(t_1\right)+v\left(t_1\right)\left(t_2-t_1\right)+(1 / 2) a\left(t_2-t_1\right)^2$
b. $v\left(t_2\right)=v\left(t_1\right)+a\left(t_2-t_1\right)$
c. $v_{\text {average }}=\left(x\left(t_2\right)-x\left(t_1\right)\right) /\left(t_2-t_1\right)$
d. $\mathrm{a}_{\text {average }}=\left(\mathrm{v}\left(\mathrm{t}_2\right)-\mathrm{v}\left(\mathrm{t}_1\right)\right) /\left(\mathrm{t}_2-\mathrm{t}_1\right)$
e. $x\left(t_2\right)=x\left(t_1\right)+v_{\text {average }}\left(t_2-t_1\right)+(1 / 2) a_{\text {average }}\left(t_2-t_1\right)^2$
f. $x\left(t_2\right)-x\left(t_1\right)=$ area under the v-t curve bounded by the $t$-axis and the dotted line shown. AnswerThe correct formulae describing the motion of the particle are (c), (d) and, (f), The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.
View full question & answer→Question 53 Marks
A man walks on a straight road from his home to a market $2.5$ km away with a speed of $5 km^{ h -1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h ^{-1}$. What is the Average speed of the man over the interval of time
i. $0$ to $30$ min ,
ii. $0$ to $50$ min ,
iii. $0$ to $40$ min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Answeri. 0 to 30 min ,
Average velocity $=$ Displacement/Time $=2.5 /(1 / 2)=5 km / h$
Average speed $=$ Distance $/$ Time $=2.5 /(1 / 2)=5 km / h$
ii. 0 to 50 min
Time $=50 min=50 / 60=5 / 6 h$
Net displacement $=0$
Total distance $=2.5+2.5=5 km$
Average velocity $=$ Displacement/Time $=0$
Average speed $=$ Distance $/$ Time $=5 /(5 / 6)=6 km / h$
iii. 0 to 40 min
Speed of the man $=7.5 km / h$
Distance travelled in first $30 min=2.5 km$
Distance travelled by the man (from market to home) in the next 10 min
$=7.5 \times 10 / 60=1.25 km$
Net displacement $=2.5-1.25=1.25 km$
Total distance travelled $=2.5+1.25=3.75 km$
Average velocity $=$ Displacement/Time $=1.25 /(40 / 60)=1.875 km / h$
Average speed $=$ Distance $/$ Time $=3.75 /(40 / 60)=5.625 km / h$
View full question & answer→Question 63 Marks
A police van moving on a highway with a speed of $30km h^{–1}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of $192km h^{–1}$. If the muzzle speed of the bullet is $150m s^{–1}$, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).
AnswerSpeed of the police van, $V p=30 \mathrm{~km} / \mathrm{h}=8.33 \mathrm{~m} / \mathrm{s}$ Muzzle speed of the bullet, $\mathrm{V}_{\mathrm{b}}=150 \mathrm{~m} / \mathrm{s}$ Speed of the thief's car, $\mathrm{V}_{\mathrm{t}}=192 \mathrm{~km} / \mathrm{h}=53.33 \mathrm{~m} / \mathrm{s}$ Since the bullet is fired from a moving van, its resultant speed can be obtained as: $=150$ $+8.33=158.33 \mathrm{~m} / \mathrm{s}$ Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief's car can be obtained as: $\mathrm{V}_{\mathrm{bt}}=\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{t}}=158.33-53.33=105 \mathrm{~m} / \mathrm{s}$
View full question & answer→Question 73 Marks
Explain clearly, with examples, the distinction between: magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
AnswerThe magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.
Whereas, total path length = AB + BC It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
View full question & answer→Question 83 Marks
A player throws a ball upwards with an initial speed of $29.4m s^{–1}$. Choose the $x = 0m$ and $t = 0s$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
AnswerDuring upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
View full question & answer→Question 93 Marks
Suggest a suitable physical situation for of the following graphs.
AnswerThe given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
View full question & answer→Question 103 Marks
Explain clearly, with examples, the distinction between: Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only]
AnswerMagnitude of average velocity = Magnitude of displacement/Time interval For the given particle, Average velocity = AC/t Average speed = Total path length/Time interval = (AB + BC)/t Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line
View full question & answer→Question 113 Marks
A highway motorist travels at a constant velocity of $45 km / h ^{-1}$ in a $30 km / h ^{-1}$ zone. A motor-cyclist police officer has been watching from behind a bill board and at the same moment the speeding motorist passes the bill board, the police officer accelerates uniformly from rest to overtake her. If the acceleration of the police officer is $10 km / h ^{-2}$, how long does he take to reach the motorist?
AnswerLet the bill board be taken as the origin. Let t be the required time.
Let P represent the position where the police officer reaches the motorist.
For the motorist. (a case of uniform motion)
when $t=0, x(0)=0, v=45 km h ^{-1} x ( t )= x (0)+ vt$. or $x ( t )= vt$
$\Rightarrow 45 km h ^{-1} \times t =45 t km \ldots$ (i)
For the police officer. (a case of accelerated motion)
when $\text{t}=0,\text{x}(0),=\text{v}(0)=0,\text{a}=10\text{km/h}^2$
Now, $\text{x(t)}=\text{x}(0)+\text{v}(0)\text{t}+\frac{1}{2}\text{at}^2$
$\text{x(t)}=0+0+\frac{1}{2}\times10\times\text{t}^2$
$=5\text{t}^2\text{km}\ \dots(\text{ii})$ Comparing (i) and (ii),
we get $5t^2 = 45t t = 9$ hour.
View full question & answer→Question 123 Marks
In Exercises $3.13$ and $3.14$, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
AnswerInstantaneous velocity is given by the first derivative of distance with respect to time i.e., $V_{in} = dx/dt$ Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time. Therefore, instantaneous speed is always equal to instantaneous velocity.
View full question & answer→Question 133 Marks
The distance x travelled by a body in a straight line is directly proportional to $t^2$. Decide on the type of motion associated. If $\text{x}\propto\text{t}^3$ what change will you observe?
Answer$\text{x}\propto\text{t}^2\therefore \text{v}\propto\text{t}$ and $\text{a}\propto \text{t}^0$ So the motion is with uniform acceleration. If $\text{x}\propto \text{t}^3,$ then $\text{v}\propto \text{t}^2$ and $\text{a}\propto\text{t}.$ The motion becomes non-uniform acceleration.
View full question & answer→Question 143 Marks
Four persons K, L, M and N start from the vertices of a square of side 'a', simultaneously and move towards the neighbour in order always with the same speed of v. When and where do they meet?
AnswerAs K, L, M and N move towards the next person in order after a short time they will be at K', L', M' and N' respectively. The size to the square reduces. It indicates that they have come closer. After next short interval if they are at K", L", M" and N", the size of the square further reduces. Finally, they will follow a curvi-linear path and meet at 0, the centre of the square. 
We know, the time taken $=\frac{\text{displacement}}{\text{Velocity in the direction of displacement}}$ $\therefore \text{t}=\frac{\text{LO}}{\text{v}\cos40^\circ}$ since v $\cos 45^\circ$ is the component of the velocity of L towards the destination. $\therefore \text{t}=\frac{\text{LO}}{\Big(\frac{\text{v}}{\sqrt{2}}\Big)}$ $=\frac{\frac{\text{a}}{\sqrt{2}}}{\frac{\text{v}}{\sqrt{2}}}=\frac{\text{a}}{\text{v}}$ View full question & answer→Question 153 Marks
A jet airplane travelling at the speed of $500 km h^{–1}$ ejects its products of combustion at the speed of $1500km h^{–1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
AnswerSpeed of the jet airplane, $V_{j e t}=500 \mathrm{~km} / \mathrm{h}$ Relative speed of its products of combustion with respect to the plane, $V_{\text {smoke }}=-1500 \mathrm{~km} / \mathrm{h}$ Speed of its products of combustion with respect to the ground $=\mathrm{V}_{\text {smoke }}^{\prime}$ Relative speed of its products of combustion with respect to the airplane, $\mathrm{V}_{\text {smoke }}=\mathrm{V}_{\text {smoke }}^{\prime}-\mathrm{V}_{\text {jet }}-1500=\mathrm{V}_{\text {smoke }}^{\prime}-500 \mathrm{~V}_{\text {smoke }}^{\prime}=-$ $1000 \mathrm{~km} / \mathrm{h}$ The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.
View full question & answer→Question 163 Marks
A car moving with a speed of $50km/ h^{-1}$ can be stopped by brakes after at least 6m. What will be the minimum stopping distance, if the same car is moving at a speed of $100km/ h^{-1}$?
AnswerApproach I:$\text{u}=50\text{km/ hr}, \text{s}=0.006\text{km}.$
$\text{v}^2-\text{u}^2=2\text{as}$
$0-(50)^2=2\times\text{a}\times0.006$
$\text{a}=-\frac{2500}{2\times0.006}\text{km/ hr}^2$
When speed of the car is 100km/hr then
$0-(100)^2=2\times\Big(\frac{-2500}{2\times0.006}\Big)\times\text{s}$
$\text{s}=0.024\text{km}=24\text{m}$
So, minimum stopping distance is 24m.
Approach II:
Stopping distance $=\text{s}=\frac{\text{u}^2}{2\text{a}}$
So, for "a" remaining the same, as u becomes 2u, s becomes 4s
i.e., 4 × 6 = 24m
View full question & answer→Question 173 Marks
A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time. (Take acceleration in the backward direction as positive).
AnswerImpulsive Force is generated by the bat: If we ignore the effect of gravity just by analyzing the motion of ball in horizontal direction only, then ball moving uniformly will return back with the same speed when a bat hits it. Acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.
The variation of acceleration with time is shown in the graph.
View full question & answer→Question 183 Marks
If the initial velocity of a particle is u and collinear acceleration at any time t is $\text{a}\propto\text{t}.$ calculate the final velocity of the particle after time t.
AnswerAcceleration $\text{a}\propto \text{t}$ ⇒ a = Kt where K is a constant. $\therefore$ Equations of motion cannot be applied. $\therefore \frac{\text{dv}}{\text{dt}}=\text{Kt}$ $\Rightarrow\text{dv}=\text{Kt dt}$ Integrating, $\text{v}=\frac{\text{Kt}^2}{2}+\text{c}$ If v = 0 at t = 0 then, c = 0 $\therefore \text{v}=\frac{\text{Kt}^2}{2}$
View full question & answer→Question 193 Marks
Sketch velocity-time graph in following situations:
i. $\mathrm{v}_0>0 ; \mathrm{a}<0 \mid \mathrm{al} \rightarrow$ constant
ii. $v_0<0 ; a>0|a| \rightarrow$ is increasing uniformly.
a : acceleration, $\mathrm{v}_0$ : initial velocity
OR
- Displacement$-$time graph of any object is shown in the figure:
Draw the velocity$-$time graph for this motion.
- What does the area under acceleration$-$time graph represent?
Answer
OR
-
- Represents change in velocity.
View full question & answer→Question 203 Marks
How does the velocity-time graph for uniform motion give a geometrical way of calculating the displacement covered during a given time t?
AnswerConsider velocity-time graph for uniform motion along a straight path. The graph is a straight line parallel to the time axis as shown in following Fig.
Let $A$ and $B$ be two points on velocity-time graph corresponding to the instants $t_1$ and $t_2$. As the motion is uniform, hence, $AA _1= BB _1= v$.
$\therefore \text { Area under } v \text { - } t \text { gaph between } t_1 \text { and } t_2$
$=\text { area } A B B_1 A_1=A A_1 \times A_1 B_1=v\left(t_2-t_1\right) \text { But velocitv is defined as } v$
$=\frac{\text { Displacement }}{\text { Time }}=\frac{x_2-x_1}{t_2-t_1}$
$\therefore v\left(t_2-t_1\right)=x_2-x_1$
$\therefore \text { area } A B B_1 A_1=\left(x_2-x_1\right)$
Hence, displacement of a particle in time intervel $\left(t_2-t_1\right)$ is numerically equal to the
area under velocity-time graph between the instants $t_1$ and $t_2$. View full question & answer→Question 213 Marks
The displacement-time graph of two bodies P and Q are represented by OA and BC respectively. What is the ratio of velocities of P and Q? $\angle \text{OBC}=60^\circ$ and $\angle \text{AOC}=30^\circ.$
AnswerVelocity of P, $v_P =\tan30^\circ=\frac{1}{\sqrt{3}}$ Velocity of $Q, v_Q =-\tan 30^\circ=-\frac{1}{\sqrt{3}}$ RAtio of velocities of P and Q is 1 : -1
View full question & answer→Question 223 Marks
A train moves from one station to another in two hours time. Its speed during the motion is shown in the graph. Determine the maximum acceleration during the journey. Also calculate the distance covered during the time interval from $0.75h$ to $1\ hour$.
AnswerWe know that the slope of the velocity-time graph gives acceleration. Change in velocity in this interval ( 0.75 h to 1 hour) $=(60-20) \mathrm{km} / \mathrm{h}^{-1}=40 \mathrm{~km} / \mathrm{h}^{-1} $
$\therefore$ Acceleration in this interval $=\frac{40 \mathrm{~km} / \mathrm{h}^{-1}}{\frac{1}{4} \mathrm{~h}}=160 \mathrm{~km} / \mathrm{h}^{-2}$
Distance covered during the time interval from 0.75 h to $1 \mathrm{~h}=$ Area under the corresponding v - t graph $=\frac{1}{1}(20+60) 0.25=10 \mathrm{~km}$.
View full question & answer→Question 233 Marks
Give three important characteristics of displacement.
AnswerThree important characteristics of displacement are:
- Displacement is a vector quantity having both magnitude as well as direction.
- Displacement of a particle between two given positions is unique and is the shortest path through which particle may go from its initial to final position.
- Displacement is independent of the choice of origin to the co$-$ordinate system.
View full question & answer→Question 243 Marks
The minute hand of a wall clock is 10cm long. Find its displacement and the distance covered from 12:00 noon to 12:30 p.m.
AnswerLength of minute hand = radius of circle described by minute hand r = 10cm = 0.1m. 
From 12:00 noon to 12:30 p.m., the tip of minute hand covers a net displacement equal to the diameter of circle. Hence, Displacement AOB = 2 × r = 2 × 0.1m = 0.2m During the same time total distance covered by tip of minute hand = semicircular path ACB $=\pi=3.14\times0.1=0.31\text{m}$ View full question & answer→Question 253 Marks
A player throws a ball upwards with an initial speed of $29.4m s^{–1}$. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take $g = 9.8m^{s–2}$ and neglect air resistance).
AnswerInitial velocity of the ball, $u=29.4 \mathrm{~m} / \mathrm{s}$ Final velocity of the ball, $v=0$ (At maximum height, the velocity of the ball becomes zero) Acceleration, $\mathrm{a}=-\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^2$ From third equation of motion, height ( s ) can be calculated as: $\mathrm{v}^2-\mathrm{u}^2$ $=2 \mathrm{gs} s=\left(\mathrm{v}^2-\mathrm{u}^2\right) / 2 \mathrm{~g}=\left((0)^2-(29.4)^2\right) / 2 \times(-9.8)=3 \mathrm{~s}$ Time of ascent $=$ Time of descent Hence, the total time taken by the ball to return to the player's hands $=3+3=6 \mathrm{~s}$.
View full question & answer→Question 263 Marks
Two balls of different masses are thrown vertically upwards with same initial speed. Which one will rise to the greater height? Which of the two will come back with greater speed to the point of projection?
AnswerSince they are thrown with same initial speed, and the equations of motion are independent of the masses, both of them will rise to the same height and will have same speed while coming back to the point of projection.
View full question & answer→Question 273 Marks
A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 - t); 0 < t < 3 and v(t) = -(t - 3)(6 - t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20m. How many cycles (counting fractions) are required to reach the top?
AnswerGiven velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ Distance covered in 0 to 3s = 9m Distance covered in 3 to 6s $=\int_3^6(18-9\text{t}+\text{t}^2)\text{dt}=\Big(18\text{t}-\frac{9\text{t}^2}{2}+\frac{\text{t}^3}{3}\Big)^6$ $=18\times6-\frac{9}{2}\times6^2+\frac{6^3}{3}-\Big(18\times3-\frac{9\times3^2}{2}+\frac{3^3}{3}\Big)$ $=108-9\times18+\frac{6^3}{3}-18\times3+\frac{9}{2}\times9-\frac{27}{3}$ $=-4.5\text{m}$ $\therefore$ Total distance travelled in one cycle $=\text{s}_1+\text{s}_2=9-4.5=4.5\text{m}$ Number of cycles to be covered in total distance $=\frac{20}{4.5}\approx4.44\approx5$
View full question & answer→Question 283 Marks
An electron travelling with a speed of $5 \times 10^3ms^{-1}$ passes through an electric field with an acceleration of $10^{12}ms^{-2}$. How long will it take for the electron to double its speed?
Answer$\text{v}(0)=5\times10^3\text{ms}^{-1},\text{a}=10^{12}\text{ms}^{-2}$
$\text{v(t)}=2\times\text{v}(0)=2\times5\times10^3\text{ms}^{-1},\text{t}=?$
Now, $\text{v(t)}=\text{v}(0)+\text{at}$ $\therefore 2\times5\times10^3=5\times10^3+10^{12}\times\text{t}$
$\Rightarrow 10^{12}\text{t}=5\times10^3$
$\Rightarrow\text{t}=\frac{5\times10^3}{10^{12}}$
$\text{t}=5\times10^{-9}\text{s}$
View full question & answer→Question 293 Marks
What is the position at any time, for a body starting from rest, with an acceleration $a = at^2$?
Answer$\text{a}=\alpha\text{t}^2$ is a variable acceleration.
$\therefore \text{dv}=\alpha\text{t}^2\text{dt}$ $|\text{v}|^\text{v}_0=\Big|\frac{\alpha\text{t}^3}{3}\Big|^\text{t}_0$
$\Rightarrow \text{v}=\frac{\alpha\text{t}^3}{3}$ Also $\text{v}=\frac{\text{dx}}{\text{dt}}$
$\therefore \text{dx}=\frac{\alpha\text{t}^3}{3}\text{dt}$
$|\text{x}|^{{\text{x}}_\text{f}}_{\text{x}_\text{i}}=\frac{\alpha\text{t}^4}{12}$
$\therefore \text{x}_\text{f}-\text{x}_\text{i}=\frac{\alpha\text{t}^4}{12}$
$\text{x}_\text{f}=\text{x}_\text{i}+\frac{\alpha\text{t}^4}{12}$ is the final position at time 't' seconds.
View full question & answer→Question 303 Marks
A body is moving with a uniform acceleration. Its velocity after 5 seconds is $25m/s$ and after $8$ seconds is $34m/s$. Calculate the distance it will cover in $10^{th}$ second.
AnswerSuppose v(0) and 'a' be the initial velocity and acceleration of the body respectively.Case I:
$\text{v(t)}=25\text{m/s}; \text{ t}=5\text{s}$
$\therefore \text{v(t)}=\text{v}(0)+\text{at}$ or $\text{v}(0)+5\text{a}=25$Case II:
$\text{v(t)}=34\text{m/s}; \text{ t}=8\text{s}$
$\therefore \text{v}(0)+8\text{a}=34$ Solving (i) and (ii), we get $a = 3ms^{-2}$
Also, $\text{S}_{\text{n}^{\text{th}}}=\text{v}(0)+\frac{\text{a}}{2}(2\text{n}-1)$ and $\text{v}(0)=10\text{ms}^{-1}$ $\text{S}_{\text{n}^\text{th}}=10+\frac{3}{2}(10\times2-1)$
$=10+\frac{57}{2}=38.5\text{m}$
View full question & answer→Question 313 Marks
Two cars $A$ and $B$ are running at velocities of $60\ km/ hr$ and $45\ km/ hr$ respectively. Calculate the relative velocity of car $A$ if :
- They are both travelling eastwards.
- Car $A$ is travelling eastwards and car $B$ is travelling westwards.
AnswerLet us take eastern direction as positive, then the western direction is negative.
$i. v_A=+60 \mathrm{~km} / \mathrm{hr} . \mathrm{V}_{\mathrm{B}}=+45 \mathrm{~km} / \mathrm{hr}$.
The relative velocity of the car $A\ w.r.t.$ car $B=v_A-v_B$
$\therefore v_{A B}=60-45=15 \mathrm{~km} / \mathrm{hr}$.
$\therefore \mathrm{v}_{\mathrm{AB}}$ is $\mathrm{km} / \mathrm{hr}$. estward.
$ii. \mathrm{v}_{\mathrm{A}}=+60 \mathrm{~km} / \mathrm{hr}$
$v_B=-45 \mathrm{~km} / \mathrm{hr}$
$v_{A B}=v_A-v_B$
$=60-(-45)$
$=105 \mathrm{~km} / \mathrm{hr}$ eastwards.
View full question & answer→Question 323 Marks
The velocity-time graph of a particle in one-dimensional motion is shown in:
Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$:
a. $x\left(t_2\right)=x\left(t_1\right)+v\left(t_1\right)\left(t_2-t_1\right)+(1 / 2) a\left(t_2-t_1\right)^2$
b. $v\left(t_2\right)=v\left(t_1\right)+a\left(t_2-t_1\right)$
c. $v_{\text {average }}=\left(x\left(t_2\right)-x\left(t_1\right)\right) /\left(t_2-t_1\right)$
d. $a_{\text {average }}=\left(v\left(t_2\right)-v\left(t_1\right)\right) /\left(\mathrm{t}_2-\mathrm{t}_1\right)$
e. $x\left(t_2\right)=x\left(t_1\right)+v_{\text {average }}\left(t_2-t_1\right)+(1 / 2) a_{\text {average }}\left(t_2-t_1\right)^2$
f. $x\left(t_2\right)-x\left(t_1\right)=$ area under the $v$-t curve bounded by the $t$-axis and the dotted line shown. AnswerThe correct formulae describing the motion of the particle are (c), (d) and, (f), The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.
View full question & answer→Question 333 Marks
A body starts accelerating uniformly with a from a velocity ‘u' and travels in a straight line. Prove that it covers a length of $\text{u}+\frac{\text{a}}{2}(2\text{t}-1)$ in the $t^{th}$ second of motion.
AnswerInitial velocity = u and acceleration = a. Length covered in t seconds $=\text{ut}+\frac{1}{2}\text{at}^2\ \dots\text{(i)}$ $=\text{u}(\text{t}-1)+\frac{1}{2}\text{a}(\text{t}-1)^2\ \dots{(\text{ii})}$
Subtracting (i) and (ii), we get Length covered in $t^{th}$ second $=\text{u}+\frac{\text{a}}{2}(2\text{t}-1).$
View full question & answer→Question 343 Marks
Identify the types of motion in following cases:
AnswerA - Body is at rest. B - Body has uniform motion-velocity is constant. C - Uniform motion but -ve velocity. D - Not a possible case, since x takes many values for same time. E - Accelerated motion, since slope (velocity) increases. F - Decelerated motion, since velocity decreases.
View full question & answer→Question 353 Marks
Give examples of a one-dimensional motion where: The particle moving along positive x-direction comes to rest periodically and moves forward.
AnswerThe equation which contains sine and cosine functions is periodic in nature. The particle will be moving along positive x-direction only if $\text{t}>\sin\text{t}$ We have displacement as a function of time, $(\text{x})=\text{t}-\sin\text{t}$ By differentiating this equation w.r.t. time we get velocity of the particle as a function of time. Velocity $\text{v}(\text{t})=\frac{\text{dx(t)}}{\text{dt}}=1-\cos\text{t}$ If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time. acceleration $\text{a}(\text{t})=\frac{\text{dv}}{\text{dt}}=\sin\text{t}$ When t = 0; x(t) = 0 When $\text{t}=\pi;\ \text{x}(\text{t})=\pi>0$ When $\text{t}=0;\ \text{x}(\text{t})=2\pi>0$
View full question & answer→Question 363 Marks
Find the displacement and distance travelled by a body in 10 seconds, using the v - t graph given below:
AnswerArea below v - t graph gives idea of distance travelled. +ve displacement $=\frac{1}{2}\times6\times5=15\text{m}$ -ve displacement $=\frac{1}{2}\times5\times4=10\text{m}$ Net displacement $=15-10=5\text{m}$ Distance travelled = 25m
View full question & answer→Question 373 Marks
Figure shows the $x$ coordinate of a particle as a function of time. Find the signs of $v_x$ and $a_x$ at $t = t_1, t = t_2$ and $t = t_3$.
Answer
- As slope is positive velocity is positive.
As slope is increasing acceleration is positive.
- As slope is zero velocity is zero.
As slope is decreasing acceleration is negative.
- As slope is negative velocity is negative.
As slope is increasing acceleration is positive.
View full question & answer→Question 383 Marks
A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 km^{ h -1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h ^{-1}$. What is the Average speed of the man over the interval of time
i. $0$ to $30$ min ,
ii. $0$ to $50$ min ,
iii. $0$ to $40$ min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Answer
i. 0 to 30 min ,
Average velocity $=$ Displacement/Time $=2.5 /(1 / 2)=5 km / h$
Average speed $=$ Distance $/$ Time $=2.5 /(1 / 2)=5 km / h$
ii. 0 to 50 min
Time $=50 min=50 / 60=5 / 6 h$
Net displacement $=0$
Total distance $=2.5+2.5=5 km$
Average velocity $=$ Displacement/Time $=0$
Average speed $=$ Distance $/$ Time $=5 /(5 / 6)=6 km / h$
iii. 0 to 40 min
Speed of the man $=7.5 km / h$
Distance travelled in first $30 min=2.5 km$
Distance travelled by the man (from market to home) in the next 10 min $=7.5 \times 10 / 60=1.25 km$
Net displacement $=2.5-1.25=1.25 km$
Total distance travelled $=2.5+1.25=3.75 km$
Average velocity $=$ Displacement $/$ Time $=1.25 /(40 / 60)=1.875 km / h$
Average speed $=$ Distance $/$ Time $=3.75 /(40 / 60)=5.625 km / h$
View full question & answer→Question 393 Marks
Two parallel rail tracks run north-south. Train A moves due north with a speed of $54km/ h^{-1}$ and train B moves due south with a speed of $90km/ h^{-1}$. What is the relative velocity of B with respect to A in $m s^{-1}$?
Answer$\mathrm{V}_{\mathrm{A}}=54 \mathrm{~km} / \mathrm{hr} \mathrm{V}_{\mathrm{B}}=90 \mathrm{~km} / \mathrm{hr}$ let due north direction be taken as + ve direction $\mathrm{V}_{\mathrm{BA}}=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-90-54=-144 \mathrm{~km} /$ $\mathrm{hr}=-40 \mathrm{~m} / \mathrm{s}$ due to south.
View full question & answer→Question 403 Marks
State which of the following situations are possible and give an example for each of these?
- An object with a constant acceleration but with zero velocity.
- An object moving in a certain direction with acceleration in the perpendicular direction.
AnswerBoth the situations are possible:
- When an object is projected upwards, its velocity at the top-most point is zero even though the acceleration on it is $9.8m/s^2(g)$.
- When a stone tied to a string is whirled in a circular path, the acceleration acting on it is always at right angles i.e. perpendicular to the direction of motion of stone $($we will study about it in chapter ‘motion in a plane'$)$.
View full question & answer→Question 413 Marks
Two trains A and B of length 400m each are moving on two parallel tracks with a uniform speed of $72km/ h^{-1}$ in the same direction, with A ahead of B. The driver of B desires to overtake A and accelerates by $1ms^{-2}$. If, after 50s, the guard of B just brushes past driver of A, calculate the original distance between the two trains.
AnswerOriginally, both the trains have the same velocities. So, the relative velocity of B w.r.t. A is zero. Now, for train B, $\text{v}(0)=0,\text{a}=1\text{ms}^{-2}$
$\text{t}=50\text{s},\text{x(t)}-\text{x}(0)=?$
$\text{x(t)}=\text{v}(0)+\text{v}(0)\text{t}+\frac{1}{2}\text{at}^2$
$\text{x(t)}-\text{x}(0)=(0)\text{t}+\frac{1}{2}\text{at}^2$
$=0\times50+\frac{1}{2}\times1\times50\times50$ $=1250\text{m}$
View full question & answer→Question 423 Marks
A car travels first half of a length S with velocity $v_1$. The second half is covered with velocities $v_2$ and $v_3$ for equal time intervals. Find the average velocity of the motion.
AnswerAverage velocity, v $=\frac{\text{Total displacement}}{\text{Total time taken}}$ Time taken to cover first half of the length $=\frac{\text{S}}{2\text{v}_1}$
Let time taken to cover second half of the length = 2t Thus, $\text{v}=\frac{\text{S}}{\frac{\text{S}}{2\text{v}_1}+2\text{t}}$ Second is divided qually into two parts with equal time.
$\therefore \frac{\text{S}}{2}=\text{v}_2\text{t}+\text{v}_3\text{t}=(\text{v}_2+\text{v}_3)\text{t}$ $2\text{t}=\frac{\text{S}}{(\text{v}_2+\text{v}_3)}$
Now, $\text{v}=\frac{\text{S}}{\frac{\text{S}}{2\text{v}_1}+\frac{\text{S}}{(\text{v}_2+\text{v}_3)}}$ $\Rightarrow \text{v}=\frac{2\text{v}_1(\text{v}_2+\text{v}_3)}{(\text{v}_2+\text{v}_3+2\text{v}_1)}$
View full question & answer→Question 433 Marks
A particle moves in a straight line such that its displacement at any time is given by $s^2 = t^2 +1$. Find
- Velocity.
- Acceleration as a function of $s$.
Answer
- $\text{s}^2=\text{t}^2+1$
Differentiating with respect to time, we get
$2\text{s}\frac{\text{ds}}{\text{dt}}=2\text{t}$
$\text{sv}=\text{t}$ or $\text{v}=\frac{\text{t}}{\text{s}}$
- Differentiating again, with respect to time, we get
$\text{s}.\text{a}+\text{v}.\text{v}=1$
$\text{a}=\frac{1-\text{v}^2}{\text{s}}=\frac{1}{\text{s}}-\frac{\text{v}^2}{\text{s}}$
$=\frac{1}{\text{s}}-\frac{\text{t}^2}{\text{s}^3}$
$=\frac{\text{s}^2-\text{t}^2}{\text{s}^3}=\frac{1}{\text{s}^3}$
$\text{a}=\frac{1}{\text{s}^3}$ View full question & answer→Question 443 Marks
Differentiate between average and instantaneous velocity.
AnswerAverage velocity: Average velocity is the displacement divided by the time interval in which the displacement occurs. $\vec{\text{v}}_{\text{av}}=\frac{\Delta\vec{\text{x}}}{\Delta\text{t}}$ Instantaneous velocity: Instantaneous velocity is defined as the limit of the average velocity as the time interval $\Delta\text{t}$ becomes infinitesimally small. $\vec{\text{v}}=\lim\limits_{\Delta\text{t} \rightarrow 0}\frac{\Delta\vec{\text{x}}}{\Delta\text{t}}=\frac{\text{dx}}{\text{dt}}$
View full question & answer→Question 453 Marks
If in a case of motion, displacement is directly proportional to the square of the time elapsed, what do you think about its acceleration i.e. constant or variable? Explain. $OR$
An object is in uniform motion, along a straight line. What will be the position time graph for the motion of the object, when
i. $x_0=+v e, v=-v e(v) \rightarrow$ constant
ii. $\mathrm{x}_0=-\mathrm{ve}, \mathrm{v}=+\mathrm{ve}(\mathrm{v}) \rightarrow$ constant
AnswerIn case of motion, displacement $(x)$ is directly proportional to the square of the time elapsed $(t)$, i.e. $\text{x}\propto \text{t}^2$
$\Rightarrow \text{x}=\text{Ct}^2$ Velocity $\text{K}=\frac{\text{dx}}{\text{dt}}$
$=\text{C}\times2\text{t} ($On differentiating $w.r.t.)$ Acceleration $\text{A}=\frac{\text{dK}}{\text{dt}}=2\text{C}=$ Constant Thus acceleration remains constant when displacement is proportional to square of time period. $OR$
- $x_0 =$ positive, $v =$ negative $\rightarrow$ constant
- $X_0 = -ve, v = +ve(v) \rightarrow$ constant
View full question & answer→Question 463 Marks
Establish the kinematic equation $v^2 - u^2 = 2as$ from velocity-time graph for a uniformly accelerated motion.
Answer
The velocity-time graph for uniformly accelerated motion has been shown in Fig. with initial velocity at t = 0 as u and final velocity at time t as v. Then area under the v-t graph gives the value of total displacement in the given time. Hence, displacement of moving particle in time t = area of trapezium $\Delta \text{ABC}$
$\text{s}=\frac{1}{2}(\text{OA}+\text{CB})\times\text{OC}$
$=\frac{1}{2}(\text{u}+\text{v})\times\text{t}$ However, from definition of acceleratiory we know that $\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$ or $\text{t}=\frac{\text{v}-\text{u}}{\text{a}}$ Substituting this value of time t in equation (i), we get Displacemen $\text{s}=\frac{1}{2}(\text{u}+\text{v})\times\frac{\text{v}-\text{u}}{\text{a}}$
$\frac{(\text{v}^2-\text{u}^2)}{2\text{a}}$
$\Rightarrow 2\text{as}=\text{v}^2-\text{u}^2$
$\text{v}^2=\text{u}^2+2\text{as}$ View full question & answer→Question 473 Marks
A motor boat covers the distance between two spots on the river in time of 8hrs. and 12hrs. downstream and upstream respectively. What is the time required for the boat to cover this distance in still water?
AnswerTime taken in downstream, $8=\frac{\text{S}}{\text{v}_\text{r}+\text{v}_\text{b}}$ Time taken for upstream, $12=\frac{\text{S}}{\text{v}_\text{b}-\text{v}_\text{r}}$ We have, $\text{v}_\text{r}+\text{v}_\text{b}=\frac{\text{S}}{8},\text{v}_\text{b}-\text{v}_\text{r}=\frac{8}{12}$ Solving, $\text{v}_\text{b}=\frac{\text{S}}{2}\Big(\frac{1}{8}+\frac{1}{12}\Big)$ $=\frac{\text{S}}{2}\Big(\frac{20}{96}\Big)=\frac{10\text{S}}{96}$ $\text{v}_\text{r}=\frac{10\text{S}}{96}-\frac{\text{S}}{12}$ $=\frac{2\text{S}}{96}$ In still water, only the velocity of boat is to be considered. $\therefore$ Time taken in still water for covering length S is, $\text{t}=\frac{\text{S}}{\text{v}_\text{b}}=\frac{\text{S}\times96}{10\text{S}}$ $=9.6\text{hrs.}$
View full question & answer→Question 483 Marks
A 100m sprinter uniformly increases his speed from rest at the rate of $1ms^{-2}$ up to $\frac{3}{4}\text{th}$ of the total run and then covers the balance $\frac{1}{4}\text{th}$ run with uniform speed. How much time does he take to complete the race?
AnswerHere total distance covered s = 100m, u = 0 For first $\frac{3}{4}\text{th}$ of the run
i.e., $\text{s}_1=\frac{3}{4}\text{s}=\frac{3}{4}\times100\text{m}=75\text{m},$
$a = +1 ms^{-2}$. If time for this part of run be $t_1$,
then using the equation s $=\text{ut}+\frac{1}{2}\text{at}^2,$
we have $75=0+\frac{1}{2}\times1\times\text{t}^2_1$
$\text{t}^2_1=75\times2=150$
$\Rightarrow\text{t}_1=\sqrt{150}=12.25\text{s}$ and .
final velocity $v = u + at_1 = 0 + 1 \times 12.25 = 12.25ms^{-1}$.
For remaining run i.e., $s_2 = s - s_1 = 100 - 75 = 25m$,
uniform velocity $v = 12.25ms^{-1}$
$\therefore$ Time for this run $\text{t}_2=\frac{\text{s}_2}{\text{v}}=\frac{25}{12.25}=2.04\text{s}$
$\therefore $ Total time taken by the sprinter to complere the race $t = t_1 + t_2 = 12.25s + 2.04s = 14.29s$
View full question & answer→Question 493 Marks
Velocity-time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement-time graph for the motion of the object.
AnswerAcceleration = zero. x-t graph is as shown below.
View full question & answer→Question 503 Marks
A woman starts from her home at 8.00 a.m., walks with a speed of 5km/ hr on a straight road upto her office 5km away stays at the office upto 4 p.m., and returns home by an auto with a speed of 25km/ hr. Choose suitable scales, and plot the x-t graph of her motion.
AnswerThe time taken by the woman to reach office is one hour. The time taken by the woman to return home from office $\frac{5}{25}=12\text{ minutes.}$ 
View full question & answer→Question 513 Marks
It is a common observation that rain clouds can be at about a kilometre altitude above the ground. If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. ($g = 10m/s^2$)
AnswerKey concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. According to the problem (h) = 1km = 1000m and we know that the initial velocity of the ball is zero. And displacement covered by rain drop in downward direction, so we will taking h as negative. (We are neglecting the air resistance.)
Velocity attaind by the rain drop in falling through a height h is $\text{v}^2=\text{u}^2-2\text{g}(-\text{h})$ As u = 0 So, $\text{v}=\sqrt{2\text{gh}}=\sqrt{2\times10\times1000}=100\sqrt{2}\text{m/s}$ $=100\sqrt{2}\times\frac{60\times60}{1000}\text{km/h}=360\sqrt{2}\text{km/h}=510\text{km/h}$ View full question & answer→Question 523 Marks
A police van moving on a highway with a speed of $30 \mathrm{~km} \mathrm{~h}^{-1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 \mathrm{~km} \mathrm{~h}^{-1}$. If the muzzle speed of the bullet is $150 \mathrm{~m} \mathrm{~s}^{-1}$, with what speed does the bullet hit the thief's car? (Note: Obtain that speed which is relevant for damaging the thief's car).
AnswerSpeed of the police van, $\mathrm{V}_{\mathrm{p}}=30 \mathrm{~km} / \mathrm{h}=8.33 \mathrm{~m} / \mathrm{s}$ Muzzle speed of the bullet, $\mathrm{V}_{\mathrm{b}}=150 \mathrm{~m} / \mathrm{s}$ Speed of the thief's car, $\mathrm{V}_{\mathrm{t}}=192 \mathrm{~km} / \mathrm{h}=53.33 \mathrm{~m} / \mathrm{s}$ Since the bullet is fired from a moving van, its resultant speed can be obtained as: $=150$ $+8.33=158.33 \mathrm{~m} / \mathrm{s}$ Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief's car can be obtained as: $\mathrm{V}_{\mathrm{bt}}=\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{t}}=158.33-53.33=105 \mathrm{~m} / \mathrm{s}$
View full question & answer→Question 533 Marks
$i.$ Explain clearly with examples the distinction between magnitude of displacement over an interval of time and the total length of the path covered by a particle over the same interval.
$ii.$ A body starting from rest accelerates uniformly along a straight line at the rate of $10\ ms^{-2}$ for $5 s$ . It moves for $2 s$ with uniform velocity of $50\ ms^{-1}$. Then it retards uniformly and comes to rest in $3 s$ . Draw velocity$-$time graph of the body and find the total distance travelled by it.
Answer$i.$ Magnitude of displacement of a particle in motion force given time is the shortest distance between the initial and final positions, while total length of path or $($path length$)$ is the length of actual path traversed by the particle in the given time.
Suppose an object goes from $A$ to $C$ following the path $A B C$, in a certain time $t$
Total length of path $=A B+B C$
When an object goes on the path $A B C$, then the displacement of the object is $(\overrightarrow{A C})$. The arrow head at $A C$ shows that the object is displaced from $A$ to $C$ .
- Given: $a = 10ms^{-2}, u = 0$
$t = 5$
$\therefore$ $v = 0 + 10 \times 5$
$v = 50 = ms^{-1}$
Area below $v-t $ graph gives distance travelled in the straight line
$\therefore$ Distance $=\frac{1}{2}(50)\times(10+2)=300\text{m}$
View full question & answer→Question 543 Marks
The displacement $($in metre$)$ of a particle moving along $x-$axis is given by $x = 18t + 5r^2$. Calculate:
- The instantaneous velocity at $t = 2s$.
- Average velocity between $t = 2s$ and $t = 3s$.
- Instantaneous acceleration.
Answer$\text{x}=18\text{t}+5\text{t}^2$
- $\text{v}=\frac{\text{dx}}{\text{dt}}=18+10\text{t}$
At $\text{t}=2\text{s, v}=18+10\times2$
$=38\ \text{ms}^{-1}$
- $\text{a}=\frac{\text{dv}}{\text{dt}}=10 ($a constant$)$
$\therefore \text{v}_{\text{av}}=\frac{\text{v}_\text{i}+\text{v}_\text{f}}{2}$
i.e., $\text{v}_\text{av}=\frac{\text{v}_{\text{t}=2}+\text{v}_{\text{t}=3}}{2}$
$=\frac{38+48}{2}=43\ \text{ms}^{-1}$
- Acceleration = $10\ ms^{-2}$.
View full question & answer→Question 553 Marks
A particle executes the motion described by $\text{x}(\text{t})=\text{x}_0(1-\text{e}^{-\gamma\text{t}});\text{t}\ge0,\text{x}_0>0.$ Where does the particle start and with what velocity?
Answer$\text{x}(\text{t})=\text{x}_0[1-\text{e}^{-\gamma\text{t}}]\ \ \ ...(\text{i})$ $\text{v}(\text{t})=\frac{\text{dx(t)}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{x}_0(1-\text{e}^{\gamma\text{t}})]=+\text{x}_0\gamma\text{e}^{-\gamma\text{t}}\ \ \ ...(\text{ii})$ $\text{a}(\text{t})=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}[+\text{x}_0\gamma^2\text{e}^{-\gamma^4}]}{2}=-\text{x}_0\gamma^2\text{e}^{-\gamma\text{t}}\ \ \ ..(\text{iii})$ $(\text{i})\text{At, t}=0\ \ \text{x}(0)=\text{x}_0[1-\text{e}^0]=\text{x}_0(1-1)=0$ $\text{v}(0)=\text{x}_0\gamma\text{e}^0=\text{x}_0\gamma$ Hence, the particle start from x = 0 with velocity $\text{v}_0=\text{x}_0\gamma.$
View full question & answer→Question 563 Marks
A bus starts with a constant acceleration $1\ ms^{-2}$. At the same time a car moving with a constant velocity of $5\ ms^{-1}$ overtakes the bus.
- How far from the starting point, the bus overtakes the car.
- How fast the bus was moving at the time of overtake?
AnswerInitial velocity of bus, $u = 0$.
Let the bus overtakes the car after time $t$.
$\therefore$ Distance travelled by bus in time $t, \text{S}_\text{b}=\text{ut}+\frac{1}{2}\text{at}^2$
$=0+\frac{1}{1}\text{t}^2=\frac{\text{t}^2}{2}$ ($\because a = 1\ ms^{-2}$)
Distance travelled by car moving with constant velocity $(a = 0)$, $\text{S}_\text{c}=\text{ut}+\frac{1}{2}\text{at}^2=5\text{t}$ $(\because u = 5\ ms^{-1}$ and $a = 0)$ Since $\text{S}_\text{b}=\text{S}_c$
$\therefore \frac{\text{t}^2}{2}=5\text{t}$
$\text{t}=10\text{s}$
$\therefore$ Distance travelled by bus when it overtakes car, $\text{S}_\text{b}=\text{ut}+\frac{1}{2}\text{at}^2$
$=0\times10+\frac{1}{2}\times(10)^2\text{s}=50\text{m}.$
Speed of bus, $v = u + at = 0 + 1 \times 10 = 10\ ms^{-1}$.
View full question & answer→Question 573 Marks
A body, starts from rest and accelerates uniformly, find the ratio of the displacement in :
- One, two and three seconds.
- First, second and third second.
Answer
- Length covered in $t$ seconds $=\frac{1}{2}\text{at}^2$
Since initial velocity is zero.
$\therefore$ Ratio of length covered in one, two and three seconds is $1 : 4 : 9$.
- Length covered in $t^{th}$ second $=\frac{\text{a}}{2}(2\text{t}-1)$
$\because\text{u}=0$
$\therefore$ Ratio of length covered in first, second and third second is $1 : 3 : 5.$ View full question & answer→Question 583 Marks
A car starting from rest, accelerates at the rate f through a distances, then continues at constant speed for sometimet and then decelerate at the rate $\frac{\text{f}}{2}$ to come to rest. If the total distance is $5s$, then prove that $\text{s}=\frac{1}{2}\text{ft}^2.$
AnswerFor accelerated motion, $\text{u}=0, \text{a}=\text{f}, \text{s}=\text{s}$ As $\text{v}^2-\text{u}^2=2\text{as},$
$\therefore \text{v}^2_1-0^2=2\text{fs}$
$\Rightarrow\text{v}_1=\sqrt{2\text{fs}},$ Distance travelled, $\text{s}_2=\text{v}_1\text{t}=\text{t}\sqrt{2\text{fs}}$ For decelerated motion, $\text{u}=\sqrt{2\text{fs}},\text{a}=-\frac{\text{f}}{2},\text{v}=0$ As $\text{v}^2-\text{u}^2=2\text{as}$
$\therefore 0^2-(\sqrt{2\text{fs}})^2=2\times\Big(-\frac{\text{f}}{2}\Big)\text{s}_3$ Distance travelled $s_3 = 2s$ Given, $\text{s}+\text{s}_2+\text{s}_3=5\text{s}$
$\Rightarrow \text{s}+\text{t}\sqrt{2\text{fs}}+2\text{s}=5\text{s}$
$\Rightarrow \text{t}\sqrt{2\text{fs}}=2\text{s}$
$\Rightarrow \text{s}=\frac{1}{2}\text{ft}^2$
View full question & answer→Question 593 Marks
A body is moving in a straight line along $x-$axis. Its distance from the origin is given by the equation $x = at^2 - bt^3$, where $x$ is in metre and $t$ is in second. Find :
- The average speed of the body in the interval $t = 0$ and $t = 2s$.
- its instantaneous speed at $t = 2s$.
AnswerThe given equation $x = at^2 - bt^3 If t =, x_0 = 0$ If $t = 2s, x_2 = 4a - 8b$
$ \because \Delta \text{x}=\text{x}_2-\text{x}_0$
$=4\text{a}-8\text{b}-0=4\text{a}-8\text{b}$
$\therefore$ Average speed in the given interval of time, $\text{v}_{\text{av}}=\frac{\Delta \text{x}}{\Delta \text{t}}=\frac{4\text{a}-8\text{b}}{2}$
$=2\text{a}-4\text{b}$ Instantaneous speed $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{at}^2-\text{bt}^3)$
$=2\text{at}-3\text{bt}^2$ At $t = 2s. v = 4a - 12\ bm/s$
View full question & answer→Question 603 Marks
In a car race, car A takes a time t seconds less than the car B and passes the finishing point with a velocity v more than that of the car B. If the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively, show that $\text{v}=\text{t}\sqrt{\text{a}_1\text{a}_2}.$
AnswerFor A: $a_1, t_2 - t$ and $v_2 + v$ be the acceleration, time taken and final velocity. For B: $a_2, t_2$ and $v_2$ be the acceleration, time and final velocity. But length travelled is same.
$\Rightarrow\text{a}_1(\text{t}_2-\text{t})^2=\text{a}_2\text{t}_2^ 2$ Solve for $\text{t}=\text{t}_2\Big(1-\sqrt{\frac{\text{a}_2}{\text{a}_1}}\Big)$ Using, $\text{v}=\text{u}+\text{at}$ we have, $\text{v}_2+\text{v}=\text{a}_1(\text{t}_2-\text{t})$ and $\text{v}_2=\text{a}_2\text{t}_2$
$\text{v}=\text{a}_1(\text{t}_2-\text{t})-\text{a}_2\text{t}_2$
$=(\text{a}_1-\text{a}_2)\text{t}_3-\text{a}_1\text{t}$ Using value of t in $\text{v}=(\text{a}_1-\text{a}_2)\text{t}_2-\text{a}_1\text{t}$ we have, $\text{v}=\text{t}\sqrt{\text{a}_1\text{a}_2}.$
View full question & answer→Question 613 Marks
A body starting from rest accelerates uniformly along a straight line at the rate of $10ms^{-2}$? for 5 seconds. It moves for $2$ seconds with uniform velocity of $50ms^{-1}$. Then it retards uniformly and comes to rest in $3s$. Draw velocity-time graph of the body and find the total distance travelled by the body.
Answer$a = 10ms^{-2}, t = 5 sec, v = 0 + 10 \times 5 = 50m s^{-1}$ Area below v-t graph gives distance travelled in the straight line.
$\therefore$ Distance $=\frac{1}{2}(50)\times(10+2)$ $=300\text{m}$
View full question & answer→Question 623 Marks
Draw the following graphs for an object under free fall :
- Variation of acceleration with respect to time.
- Variation of velocity with respect to time.
- Variation of distance with respect to time.
Answer
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Let $2$ will be used only when the body bouinces.
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A particle located at x = 0 at time t = 0 starts moving along the positive x direction with a velocity v that varies as $\text{v}=\alpha\sqrt{\text{x}}.$ How do the displacement, velocity and acceleration of the particle vary with time? What is the average velocity of the particle over the first s metres of its path?
AnswerWe know that the instantaneous velocity of the particle is given by $\text{v}=\frac{\text{dx}}{\text{dt}}$ Since $\text{v}=\alpha\sqrt{\text{x}}$ we have $\frac{\text{dx}}{\text{dt}}=\alpha\sqrt{\text{x}}$ $\frac{\text{dx}}{\sqrt{\text{x}}}=\alpha\text{ dt}$ Integrating from t = 0 (x = 0) to t = t (x = x) we have $\int \limits^\text{x}_0\text{x}^{\frac{-1}{2}}\text{dxs}=\alpha\int\limits^\text{t}_0\text{dt}$ $\therefore \Bigg|\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}\Bigg|^\text{x}_0=\alpha\text{ t}$ $\text{x}=\frac{\alpha^2\text{t}^2}{4}$ The time dependence of the velocity is obtained by differentiating both sides of this relation w.r.t. time t. Thus $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\alpha.2\text{ t}}{4}$ $=\frac{\alpha^2}{2}\text{t}$ The velocity x of the particle is thus increasing in direct proportion to time. Similarly the time dependence of acceleration is obtained by differentiating both sides of this relation w.r.t.'t'. Thus $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\alpha^2}{2}$ The particle is thus moving with a constant acceleration. To find the average velocity over the first s metre, we assume that the time taken to cover this distance is T. Using $\text{x}=\frac{\alpha^2\text{t}^2}{4}$ we get $\text{s}=\frac{\alpha^2\text{T}^2}{4}$ $\text{T}=\frac{2\sqrt{\text{s}}}{\alpha}$ The average velocity $\text{v}_{\text{av}}\Big(=\frac{\text{s}}{\text{T}}\Big)$ is, therefore $\text{v}_{\text{av}}=\Big(\frac{\alpha}{2}\sqrt{\text{s}}\Big)$
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If a body travels half its total path in its last second of its fall from rest, calculate the time and height of its fall.
AnswerT = total time taken by the body to fall down h = total height of the fall Using $\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2,$ we have $\text{h}=0+\frac{1}{2}\text{gt}^2=\frac{1}{2}\text{at}^2\ \dots(\text{i})$ Now, $\text{t}=(\text{T}-1),\text{h}=\frac{\text{h}}{2}$ $\therefore \frac{\text{h}}{2}=0+\frac{1}{2}\text{g}(\text{T}-1)^2$ $\text{h}=\text{g}(\text{T}-1)^2\ \dots(\text{ii})$ From equations (i) and (ii), we have $\frac{1}{2}\text{gT}^2=\text{g}(\text{T}-1)^2$ $\text{T}^2=2(\text{T}-1)^2$ $\text{T}^2-4\text{T}+2=0$ $=\therefore \text{T}\pm1.141$ $\text{T}=3.414\text{s}$ or $0.586\text{s}$ Since T cannot be less than 1 $\therefore \text{T}=3.414\text{s}$ From equation (i) $\text{h}=\frac{1}{2}\times9.8\times(3.414)^2=57.11\text{m}$
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A ship is moving at a speed of $56km/ h^{-1}$. One second later, it is moving at a speed of $58km/ h^{-1}$. What is its acceleration?
AnswerHere, Initial speed $u = 56km/ h^{-1} =56\times\frac{5}{18}\text{ms}^{-1}=\frac{140}{9}\text{ms}^{-1}$
$=15.55\text{ms}^{-1}$ Final speed, $v = 58km/ h^{-1} =58\times\frac{5}{18}\text{ms}^{-1}=\frac{145}{9}\text{ms}^{-1}$ $=16.11\text{ms}^{-1}$ Time taken = 1s Using equation of motion $\text{v}=\text{u}+\text{at}$
$\Rightarrow \text{a}=\frac{\text{v}-\text{u}}{\text{t}}$
$\Rightarrow \text{a}=\Big(\frac{16.11-15.55}{1}\Big)=0.56\text{ms}^{-2},$
$\text{a}=0.56\text{ms}^{-2}$
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The momentum p of a particle changes with time t according to the relation $\frac{\text{dp}}{\text{dt}}=(10\text{N})+(2\text{N/s)t}.$ If the momentum is zero at t = 0, what will the momentum be at t = 10s?
Answer$\frac{\text{dp}}{\text{dt}}=(10\text{N})+(2\text{N/s)t}$ momentum is zero at t = 0 $\therefore$ momentum at t = 10 sec will be dp = [(10 N) + 2Ns t]dt $\int\limits^{\text{P}}_0\text{dp}=\int\limits^{10}_010\text{dt}+\int\limits^{10}_0(2\text{tdt})=10\text{t}]^{10}_0+2\frac{\text{t}^2}{2}\Big]^{10}_0=200\text{kg m/s}.$
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Explain clearly, with examples, the distinction between: magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
AnswerThe magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.
Whereas, total path length = AB + BC It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
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The velocity$-$time graph of an object moving along a straight line is as shown. 
Calculate distance covered by object between:
- $t = 0$ to $t = 5 sec.$
- $t = 0$ to $t = 10 sec.$
Answer
- Distance covered by the object $t = 0$ to $t= 5 sec.$
$=$ area $\text{OABDO}$
$=\frac{1}{2} ($sum of parallel side$) \times h$
$=\frac{1}{2}(3+5)\times20$
$=4\times20=80\text{m}.$
- Distance covered by the object between
$t = 0$ to $t = 10 sec.$
$=$ area $\text{OABCO}$
$=\frac{1}{2} ($sum of parallel side$) \times h$
$=\frac{1}{2}(3+10)\times20=130\text{m}.$ View full question & answer→Question 693 Marks
A race car is moving on a straight road with a speed of $180km/ h^{-1}$. If the driver stops the car in $25s$ by applying brakes, calculate the distance covered by the car during the time brakes are applied. Assume acceleration of the car be uniform throughout the retarding motion.
Answer$\text{u}=180\frac{\text{km}}{\text{hr}}$
$=\frac{180\times1000}{60\times60}=50\text{m/s}$
$\text{v}=\text{u}+\text{at}$
$0=50+\text{a}(25)$
$\text{a}=-2\text{m/s}^2$
$\text{v}^2-\text{u}^2=2\text{as}$
$0-(50)^2=2(-2)\times\text{s}$
$\text{s}=\frac{2500}{4}=625\text{m}$
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It is a common observation that rain clouds can be at about a kilometre altitude above the ground. Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is $5cm$. (Assume that umbrella is circular and has a diameter of 1m and cloth is not pierced through !!)
AnswerKey concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. Radius of the umbrella (R) $=\frac{1}{2}\text{m}$
$\therefore$ Area of the umbrella $(\text{A})=\pi\text{R}^2=\frac{22}{7}\times\Big(\frac{1}{2}\Big)^2$
$=\frac{22}{28}=\frac{11}{14}=0.8\text{m}^2$ Number of drops striking the umbrella simultaneously with average separation of $5cm (5 \times 10^{-2}m) \text{n}=\frac{0.8}{(5\times10^{-2})^2}=320$
$\therefore$ Net force exerted on umbrella $=320\times168=53760\text{N}=54000\text{N}$
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A motor boat covers the distance between two spots on the river in time of 8 hours and 12 hours downstream and upstream respectively. What is the time required for the boat to cover this distance in still water?
AnswerTime taken in downstream, $8=\frac{\text{S}}{\text{v}_\text{r}+\text{v}_\text{b}}$ Time taken in upstream, $12=\frac{\text{S}}{\text{v}_\text{b}-\text{v}_\text{r}}$ Given, $\text{v}_\text{r}+\text{v}_\text{b}=\frac{\text{S}}{8},$ $\text{v}_\text{b}-\text{v}_\text{r}=\frac{\text{S}}{12}$ By solving the equations, we get $\text{v}_\text{b}=\frac{\text{S}}{2}\Big(\frac{1}{8}+\frac{1}{12}\Big)$ $\text{v}_\text{b}=\frac{\text{S}}{2}\times\frac{20}{96}$ $\text{v}_\text{b}=\frac{10\text{S}}{96}$ Now, $\text{v}_\text{r}=\frac{10\text{S}}{96}-\frac{\text{S}}{12}=\frac{2\text{S}}{96}$ In still water, only the velocity is to be considered $\therefore$ time taken in still water for covering length S i.s, $\text{t}=\frac{\text{S}}{\text{v}_\text{b}}=\frac{\text{S}\times96}{10\text{S}}$ $=9.6\text{ seconds}.$
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What will be the tension in the cloth inclined at an angle of 30° from horizontal when a person of mass $50kg$ falls through it with on acceleration of $2m/s^2$.
AnswerHere $m = 50kg, \theta=30^\circ, a = 2m/s^2, g = 10m/s^2$
$\therefore$ Tension $\text{T}=\text{m}[\text{g}\sin\theta-\text{a}]$ $=50[10\times\sin30^\circ-2]$
$=50\times3$ $=150\text{N}$
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Two parallel rail tracks run north-south. Train A moves north with a speed $54 km / h ^{-1}$ and $\operatorname{train} B$ moves south with a speed of $90 km / h ^{-1}$. What is the:
i. Velocity of $B$ with respect to $A$ ?
ii. Velocity of ground with respect to B?
iii. Velocity of a monkey running on the roof of train $A$ against its motion (with a speed of $10 km^{-1}$ with respect to train A ) as observed by a man standing on the ground?
OR
Obtain any one equation of motion for constant acceleration using method of calculus.
AnswerLet the motion from north to south be taken positive. Here $\vec{\text{v}}_\text{A}=54\text{km/ h}$ (North to south) $\vec{\text{v}}_\text{B}=-90\text{km/ h}$ (South to North) 
- Velocity of B w.r.t. A is
$\vec{\text{v}}_\text{BA}=\vec{\text{v}}_\text{B}-\vec{\text{v}}_\text{A}$
$=(-90-54)$ (North to south)
$=-144\text{km/ h}$ (North → south)
$=144\text{km/ h}$ (South to north)
- Let $\vec{\text{v}}_\text{g}$ be the velocity of ground.
$\vec{\text{v}}_\text{g}=0$
Velocity of ground w.r.t. B
$\vec{\text{v}}_\text{GB}=\vec{\text{v}}_\text{G}-\vec{\text{v}}_\text{B}=0-(-90)$
$=90\text{km/ h}$ (North to south)
- Speed of monkey, $\text{v}_\text{C}=-10\text{km/ h},\text{v}_\text{A}=54\text{km/ h}$
$\text{v}'_\text{i}=$ speed of monkey w.r.t stationary observer
$\text{v}'_\text{i}=\text{v}_\text{C}+\text{v}_\text{A}$
$=-10+54=44\text{km/ h}$
Alternate Answer:
Velocity-time relation: Let at any instant of time t, dv be the change in velocity in time interval dt. Thus, acceleration is: $\text{a}=\frac{\text{dv}}{\text{dt}}$ $\text{dv}=\text{a dt}$ Integrating it within the condition of motion. $\int\limits^{\text{v}}_\text{u}\text{dv}=\text{a}\int\limits^{\text{t}}_\text{0}\text{dt}$ $\text{v}-\text{u}=\text{at}$ $\text{v}=\text{u}+\text{at}$ View full question & answer→Question 743 Marks
At $\mathrm{t}=0$ a particle is at rest at origin. Its acceleration is $2 \mathrm{~ms}^{-2}$ for the first $3\ s$ and $-2 \mathrm{~ms}^{-2}$ for the next $3\ s$. Plot the acceleration versus time, velocity versus time and position versus time graph.
AnswerWe are given that for first 3s acceleration is $2ms^{-2}$ and for next 3s acceleration is $2ms^{-2}$. Hence acceleration time graph is as shown in the figure.
The area enclosed between a-t curve and t-axis gives change in velocity for the corresponding interval. Also at $t = 0$, hence final velocity at $t = 3s$ will increase to $6ms^{-1}$. In next 3s the velocity will decrease to zero. Hence the velocity time-graph is as shown in figure.
Note that v-t curves are taken as straight line as acceleration is constant. Now for displacement time curve, we will use the fact that area enclosed between u-t curve and time axis gives displacement for the corresponding interval. Hence displacement in first three seconds is 9m and in next three seconds is 9m. Also the x-t curve will be of parabolic nature as motion is with constant acceleration. Therefore x-t curve is as shown in figure below.
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What causes variation in velocity of a particle?
AnswerThe velocity of a particle changes due to either of the following three causes :
- Change in magnitude of velocity.
- Change in direction of motion only.
- Change in magnitude as well as direction of the motion.
View full question & answer→Question 763 Marks
A motorboat covers the distance between two ports on the river in $t_1 = 8hr$ and $t_2 = 12hr$ downstream and upstream respectively. What is the time required for the boat to cover this distance in still water?
AnswerLet x be the distance between the two ports. Let u be the velocity of the flow of water and v be the velocity of the boat in still water.
$\therefore \text{v}+\text{u}=\frac{\text{x}}{\text{t}_1}$ $\text{v}-\text{u}=\frac{\text{x}}{\text{t}_2}$
$\therefore \text{x}\Big(\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big)=2\text{v}$
$\text{t}=\frac{\text{x}}{\text{v}}=\frac{2\text{t}_1\text{t}_2}{\text{t}_1+\text{t}_2}$
$=\frac{2\times8\times12}{8+12}=9.6\text{hr}$
View full question & answer→Question 773 Marks
Establish the kinematic equation $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$ from velocity-time graph for a uniformly accelerated motion.
AnswerLet AB be a velocity-time graph for uniformly accelerated motion with initial velocity u at time t = 0 and acceleration of the particle under motion being given by $\text{a}=\tan\theta =\frac{\text{BD}}{\text{AD}}$ We know that area under the v-t graph gives the value of displacement during that time. $\therefore$ Displacement of particle in time t will be s = area under v-t graph = area OABC = Area of rectangle OADC + area of triangle ADB $=\text{OA}\times\text{OC}+\frac{1}{2}\text{AD}\times\text{DB}\\=\text{u}\times\text{t}+\frac{1}{2}(\text{AD})\times\Big(\frac{\text{AD}\times\text{DB}}{\text{AD}}\Big)$ $=\text{ut}+\frac{1}{2}(\text{AD})^2\times\Big(\frac{\text{DB}}{\text{AD}}\Big)\text{s}=\text{ut}+\frac{1}{2}\text{t}^2\text{a}$ $\therefore \text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
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A player throws a ball upwards with an initial speed of $29.4m s^{–1}$. Choose the $x = 0m$ and $t = 0s$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
AnswerDuring upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
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A particle moving in a straight line covers half the distance with a speed of $3m/s$. The other half of the distance is covered in two equal intervals of time with speeds of $4.5m/s$ and $7.5m/s$, respectively. Find the average speed of the particle during this motion.
AnswerTime to cover $\frac{\text{s}}{2}$ distance, $\text{t}_1=\frac{\frac{\text{s}}{2}}{3}=\frac{\text{s}}{6}\text{s}$ Time to cover $s_1$ distance, $\text{t}_2=\frac{\text{s}_1}{4.5}\text{s}$ Time to cover $s_2$ distance, $\text{t}_3=\frac{\text{s}_2}{7.5}\text{s}$
Now. $\text{s}_1+\text{s}_2=\frac{\text{s}}{2}$
$\therefore 4.5\text{t}_2+7.5\text{t}_3=\frac{\text{s}}{2}$ Since, $\text{t}_2=\text{t}_3$ $\Rightarrow 4.5\text{t}_2+7.5\text{t}_2=\frac{\text{s}}{2}$
$\Rightarrow \text{t}_2=\frac{\text{s}}{24}\text{s}$
$\therefore $ Total time $=\text{t}_1+\text{t}_2+\text{t}_3=\frac{\text{s}}{6}+\frac{\text{s}}{24}+\frac{\text{s}}{24}$ $=\frac{6}{24}\text{s}=\frac{1}{4}\text{s}$
Hence, average speed $\text{s}_{\text{av}}=\frac{\text{s}}{\frac{\text{s}}{4}}=4\text{m/s}$
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Water drops fall freely from a tap at a height of $4.9m$. If time interval between successive drops is equal and the $4th$ drop is released when the first lands on the ground, find the separation between the second and third drops.
Answerh = 4.9m Time for a drop to reach ground $=\sqrt{\frac{2\times4.9}{\text{g}}}=1\text{ second}.$ when the first drop lands on ground, the $4^{th}$ is ready to come down. Therefore there are 3 equal time gaps of $\Big(\frac{1}{3}\Big)$ between two consecutive drops. Time of travel of 2nd and 3rd drops are and respectively.
Separation between $2^{nd}$ and $3^{rd}$ drops $\frac{2}{3}$ and $\frac{1}{3}$ respectively. Seperation between $2^{nd}$ and $3^{rd}$ drops $=\frac{1}{2}\text{g}\Bigg[\Big(\frac{2}{3}\Big)^2-\Big(\frac{1}{3}\Big)^2\Bigg]=4.9$ $=\Bigg[3\times\Big(\frac{1}{3}\Big)^2\Bigg]=\frac{4.9}{3}$
$=1.63\text{m}$
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A particle moves along a semicircular path of radius $R$ in time $t$ with constant speed. 
For particle calculate:
- Distance travelled.
- Displacement.
- Average speed.
- Average velocity.
- Average acceleration.
Answer
- Distance $=$ length of path of particle $=\text{AB}=\pi\text{R}$
- Displacement $=$ minimum distance between initiar and finar point $= AB = 2R$
- Average speed, $\text{v}=\frac{\text{Distance}}{\text{Time}}=\frac{\pi\text{R}}{\text{t}}$
- Average velocity $=\frac{2\text{R}}{\text{t}}\Big(\frac{\text{Displacement}}{\text{Time}}\Big)$
- Average acceleration $=\frac{\text{Change in velocity}}{\text{Time taken}}=\frac{2\text{v}}{\text{t}}$
$=\frac{2\pi\text{R}}{\text{t}^2}$ View full question & answer→Question 823 Marks
A rod of length $L$ is placed along the $X-$axis between $x = 0$ and $x = L$. The linear density $($mass$/$ length$) \rho$ of the rod varies with the distance x from the origin as $\rho=\text{a + bx.}$
- Find the $SI$ units of $a$ and $b.$
- Find the mass of the rod in terms of $a, b$ and $L.$
Answer$\rho=\frac{\text{mass}}{\text{length}}=\text{a + bx}$
- $S.I$. unit of $‘a’ = \ kg/m$ and $SI$ unit of $‘b’ = kg/m^2 ($from principle of homogeneity of dimensions$)$
- Let us consider a small element of length $‘dx\ ’$ at a distance $x$ from the origin as shown in the figure.
$\therefore\ dm =$ mass of the element $=\rho\text{ dx = (a + bx)dx}$ So, mass of the rod $= m$ $\int\text{dm}=\int\limits^{\text{L}}_0(\text{a + bx)dx}$
$=\Big[\text{ax}+\frac{\text{bx}^2}{2}\Big]^{\text{L}}_0=\text{aL}+\frac{\text{bL}^2}{2}$
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A particle starts from rest at $t = 0$ and has an acceleration as given in the figure below. Draw the v-t graph for $4$ seconds. 
AnswerVelocity at the end of 2 seconds $=0+3 \times 2=6 \mathrm{~ms}^{-1}$ Velocity at the end of 4 seconds $=6-3 \times 2=0 \mathrm{~ms}$ (Consider $t$ $=2$ to $t=4$ seconds)
View full question & answer→Question 843 Marks
Suggest a suitable physical situation for of the following graphs.
AnswerThe given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
View full question & answer→Question 853 Marks
A jet plane beginning its take off moves down the runway at a constant acceleration of $4.00\ m/s^2$.
- Find the position and velocity of the plane $5.00s$ after it begins to move.
- If a speed of $70.0\ m/s$ is required for the plane to leave the ground, how long a runway is required?
AnswerBecause the acceleration is constant, we can apply the equations of motion derived above.
- We take the origin of the $x-$axis to be the initial position of the plane, so that $x_0 = 0$. It is useful to begin by listing all the data given in the problem.
$a = 4.00\ m/s^2$
$v = 0, x = 0$
The problem may be stated in terms of the symbols as follows
Find $x$ and $v$ at $t = 5.00S$
When $x$ and $u$ are zero, these two equations reduce to $v = at$ and $\text{x}=\frac{1}{2}\text{at}^2$
At $t = 5.00s$
$\text{v}=(4.00\text{m/s}^2)(5.00\text{s})=20.0\text{m/s}$
$\text{x}=\frac{1}{2}(4.00\text{m/s}^2)(5.00\text{s})^2$
$\text{x}=50.0\text{m}$
- The problem here may be stated as
Find $x$ when $v = 70.0m/s$
It contains the single unknown $x$, as well as a and $v$, which are known with $u= 0, \text{v}^2_0=2\text{a}_{\text{x}}\text{x}$
Solving for $x$, we obtain
$\text{x}=\frac{\text{v}^2}{2\text{a}}=\frac{(70.00\text{m/s})^2}{2(4.00\text{m/s}^2)}$
$=613\text{m}$ View full question & answer→Question 863 Marks
Give examples of a one-dimensional motion where: The particle moving along positive x-direction comes to rest periodically and moves backward.
AnswerEquation can be represented by, $\text{x}(\text{t})=\sin\text{t}$ $\text{v}=\frac{\text{d}}{\text{dt}}\text{x}(\text{t})=\cos\text{t}\ \text{ and}\ \ \text{a}=\frac{\text{dv}}{\text{dt}}=-\sin\text{t}$ At t = 0; x = 0, v = 1 (positive) and a = 0 At $\text{t}=\frac{\pi}{2};\text{x}=1$ (positive), v = 0 and a = -1 (negative) At $\text{t}={\pi};\text{x}=0,\text{v}=-1$ At $\text{t}=\pi,\text{x}=0,\text{v}=-1$ (negative) and a = 0 At $\text{t}=\frac{3\pi}{2};\text{x}=-1$ (negative), v = 0 and a = +1 (positive) At $\text{t}=2\pi,\text{x}=0,\text{v}=1$ (positive) and a = 0 Hence the particle moving along positive x-direction comes to rest periodically and moves backward. As displacement and velocity is involving sin t and cos t, hence these equations represent periodic nature.
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A body covers half of its journey with a speed of 40m/s and other half with a speed of 60m/s. What is the average speed during the whole journey?
AnswerAverage speed $=\frac{\text{Total distance}}{\text{Time taken}}$ Let x be the distance to be covered $\therefore$ average speed $=\frac{\text{x}}{\frac{\text{x}}{2\text{v}_1}+\frac{\text{x}}{2\text{v}_2}}$ where, $\frac{\text{x}}{2\text{v}_1}=$ time taken to cover first half of the distance $\frac{\text{x}}{2\text{v}_2}=$ time taken to cover second half of the distance Now, average speed $=\frac{\text{x}}{\frac{\text{x}(\text{v}_1+\text{v}_2)}{2\text{v}_1\text{v}_2}}=\frac{\text{x}\times2\text{v}_1\text{v}_2}{\text{x}(\text{v}_1+\text{v}_2)}$ $=\frac{2\text{v}_1\text{v}_2}{\text{v}_1+\text{v}_2}$ $\Rightarrow \text{V}_{\text{av}}=\frac{2\times40\text{m/s}\times60\text{m/s}}{100\text{m/s}}=48\text{ms}^{-1}$
View full question & answer→Question 883 Marks
An object is thrown vertically upward with some speed. It crosses two points $p$ and $q$ which are separated by $h$ metre. If $t_p$ is the time between $p$ and the highest point and coming back and $t_q$ is the time between $q$ and the highest point and coming back, relate acceleration due to gravity, $\mathrm{t}_{\mathrm{p}}, \mathrm{t}_{\mathrm{q}}$, and h .
AnswerLet H = distance between point p and the highest point A. This distance H travelled while falling from A top is given by 
$\therefore \text{H}=\frac{1}{2}\text{g}\Big(\frac{\text{t}_\text{p}}{2}\Big)^2$
$=\frac{\text{g t}^2_\text{p}}{8}\ \dots(\text{i})$ Also, $\text{H}-\text{h}=\frac{1}{2}\text{g}\Big(\frac{\text{t}_\text{q}}{2}\Big)^2$
$=\frac{\text{g t}^2_\text{q}}{8}\ \dots(\text{ii})$
Now, From eqns. (i) and (ii), we get $\text{h}=\frac{\text{g t}^2_\text{p}}{8}-\frac{\text{g t}^2_\text{q}}{8}$ $=\frac{\text{g}}{8}(\text{t}^2_\text{p}-\text{t}^2_\text{q})$ View full question & answer→Question 893 Marks
A particle located at x = 0 at t = 0 starts moving along the positive x-direction with a velocity v that varies as $\text{v}=\alpha\sqrt{\text{x}}.$ How does the displacement of the particle vary with time?
AnswerGiven $\text{v}=\alpha\sqrt{\text{x}}$ Since $\text{v}=\frac{\text{dx}}{\text{dt}}$ $\therefore \frac{\text{dx}}{\text{dt}}=\alpha\sqrt{\text{x}}$ $\frac{\text{dx}}{\sqrt{\text{x}}}=\alpha,\text{dt}$ $\text{x}^{\frac{-1}{2}}\text{dx}=\alpha\text{dt}$ By integrating both sides, we get $\int\text{x}^{\frac{-1}{2}}\text{dx}=\int\alpha\text{dt}$ $\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}=\alpha\text{ t}$ $\text{x}^{\frac{1}{2}}=\frac{1}{2}\alpha\text{ t}$ Squaring both sides, we get $\text{x}=\frac{\alpha^2\text{t}^2}{4}$ $\therefore \text{x}\propto\text{t}^2.$
View full question & answer→Question 903 Marks
The acceleration experienced by a boat, after its engine is cut off, given by, $\frac{\text{dv}}{\text{dt}}=-\text{kv}^3,$ where k is a constant. If $v_0$ is the magnitude of velocity at cut off $(t = 0)$, find the magnitude of the velocity at a time t after the cut off.
Answer$\frac{\text{dv}}{\text{dt}}=-\text{kv}^3$ Integrating both sides, we get $\int\frac{\text{dv}}{\text{v}}=-\text{k}\int\text{dt}$ $-\frac{1}{2\text{v}^2}=-\text{kt}+\text{c}$
At $t = 0, v = v_0 \therefore \text{c}=-\frac{1}{2\text{v}^2_0}$
$\therefore -\frac{1}{2\text{v}^2}=-\text{kt}-\frac{1}{2\text{v}^2_0}$
$2\text{v}^2=\frac{2\text{v}^2_0}{(2\text{v}^2_0\text{kt}+1)}\text{s}$
$\text{v}=\sqrt{\frac{\text{v}^2_0}{(2\text{v}^2_0\text{kt}+1)}}$
View full question & answer→Question 913 Marks
A body is projected vertically upwards from A, the top of a tower it reaches the ground in it $t_1$, second. If it is projected vertically downwards from A with the same velocity it reaches the ground in $t_2$ second. If it falls freely, from A, prove that it would reach the ground in $\sqrt{\text{t}_1\text{t}_2}$ second.
AnswerUsing relatons Consider upwards as negative and downwards as positive. $\text{h}=-\text{ut}_1+\frac{1}{2}\text{gt}^2_1\ \dots{\text{i}}$ and $\text{h}=\text{ut}_2+\frac{1}{2}\text{gt}^2_2\ \dots(\text{ii})$ On subtracting Eqs. (i) from (ii), we get $0=\text{u}(\text{t}_2+\text{t}_1)+\frac{1}{2}\text{gt}^2_2-\frac{1}{2}\text{gt}^2_1$
$\text{u}(\text{t}_2+\text{t}_1)+\frac{1}{2}\text{g}(\text{t}_2+\text{t}_1)(\text{t}_2-\text{t}_1)=0$
$\text{u}+\frac{1}{2}\text{g}(\text{t}_2+\text{t}_1)=0\ \dots(\text{iii})$ From Eqs. (i) and (iii),
we get Now, $\text{h}=\frac{\text{gt}_1}{2}(\text{t}_2-\text{t}_1)+\frac{1}{2}\text{gt}^2_1=\frac{1}{2}\text{gt}_1\text{t}_2\ \dots(\text{iv})$
Again, when the body falls freely. $\text{h}=\frac{1}{2}\text{gt}^2;\frac{1}{2}\text{gt}_1\text{t}_2=\frac{1}{2}\text{gt}^2$ [from Eq. (iv)] $\text{t}=\sqrt{\text{t}_1\text{t}_2}$
View full question & answer→Question 923 Marks
A body starts from a point $A$ and moves to $B$. If it returns to the same point, find the :
- Displacement.
- Distance.
- Velocity.
- Average speed.
$[$Given: $AB = s,$ velocity $= v]$ Answer
- Displacement is zero, since it returns to $A$.
- Distance is twice the separation between $A$ and $B$.
- Velocity is zero.
- Average speed $=\frac{\text{total distance}}{\text{time}}$
$=\frac{2\text{AB}}{\frac{\text{AB}}{\text{v}}+\frac{\text{AB}}{\text{v}}}=\frac{2\text{S}}{\frac{2\text{S}}{\text{v}}}$
$=\text{v}\text{ ms}^{-1}$ View full question & answer→Question 933 Marks
A particle executes the motion described by $\text{x}(\text{t})=\text{x}_0(1-\text{e}^{-\gamma\text{t}});\text{t}\ge0,\text{x}_0>0.$ Find maximum and minimum values of x(t), v(t), a(t). Show that x(t) and a(t) increase with time and v(t) decreases with time.
Answerx(t) is minimum at $\text{t}=0\ \because\ \text{At t}=0,[\text{x}(\text{t})]_\text{min}=0$ x(t) is maximum at $\text{t}=\infty\ \because\ \text{At t}=\infty[\text{x}(\text{t})]_\text{max}=\text{e}-\gamma\text{t}=\infty$ v(t) is maximum at $\text{t}=0\ \because\ \text{At t}=0;\text{v}(0)=\text{x}_0\gamma$ v(t) is minimum at $\text{t}=\infty\ \because\ \text{At t}=\infty\text{ v}(\infty)=0$ a(t) is maximum at $\text{t}=\infty\ \because\ \text{At t}=\infty\text{ a}(\infty)=0$ a(t) is minimum at $\text{t}=0\because\text{At t}=0\ \text{a}(0)=-\text{x}_0\gamma^2$
View full question & answer→Question 943 Marks
Explain clearly, with examples, the distinction between: Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only]
AnswerMagnitude of average velocity = Magnitude of displacement/Time interval For the given particle, Average velocity = AC/t Average speed = Total path length/Time interval = (AB + BC)/t Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line
View full question & answer→Question 953 Marks
A body covers 200cm in the first 2 seconds and 220cm in the next two seconds. What will be its velocity at the end of 7 seconds? Also, find the displacement in 7 seconds?
AnswerFor first 2 seconds, $200=2\text{u}+\frac{1}{2}\text{a}\times2^2$ For filst 4 seconds, $420=4\text{u}+\frac{1}{2}\text{a}\times4^2$ Solve for u and a to get, $\text{u}=95\text{cm s}^{-1}, \text{ a}=5\text{cm s}^{-2}$ Displacement in 7 seconds $7\text{u}+\frac{1}{2}\text{a}\times7^2$ $=7\times95+\frac{1}{2}\times0.5\times7^2$ $=787.5\text{cm}$ velocity at the end of 7 seconds $=\text{v}_7=95+5\times7=130\text{cm s}^{-1}$
View full question & answer→Question 963 Marks
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $\left(v_0\right)$ and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.
AnswerLet the distance travelled by the vehicle before it stops be $d_s$. Then, using equation of motion $v^2=v_0^2+2 a x$, and noting that $v=0$, we have the stopping distance
$
d_s=\frac{-v_0^2}{2 a}
$
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).
For the car of a particular make, the braking distance was found to be $10 m , 20 m , 34 m$ and $50 m$ corresponding to velocities of $11,15,20$ and $25 m / s$ which are nearly consistent with the above formula.
Stopping distance is an important factor considered in setting speed limits, for example, in school zones.
View full question & answer→Question 973 Marks
Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely. 1: 3: 5: 7.......]" Prove it.
AnswerLet us divide the time interval of motion of an object under free fall into many equal intervals $\tau$ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have
$
y=-\frac{1}{2} g t^2
$
Using this equation, we can calculate the position of the object after different time intervals, $0, \tau, 2 \tau, 3 \tau \ldots$ which are given in second column of Table 2.2. If we take $(-1 / 2) g \tau^2$ as $y_0$ - the position coordinate after first time interval $\tau$, then third column gives the positions in the unit of $y_0$. The fourth column gives the distances traversed in successive $\tau$ s. We find that the distances are in the simple ratio $1: 3: 5: 7: 9: 11 \ldots$ as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.
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