Question 11 Mark
Shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
AnswerNo, because the x-t graph does not represent the trajectory of the path followed by a particles. From the graph, it is noted that at t = 0, x = 0.
View full question & answer→Question 21 Mark
Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
AnswerThe given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
View full question & answer→Question 31 Mark
A ball is dropped from a height of $90m$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed$-$time graph of its motion between $t = 0$ to $12s$.
Answer
- Downward motion of Ball
Ball is dropped from a height, $s=90 \mathrm{~m}$
Initial velocity of the ball, $\mathrm{u}=0$
Acceleration, $a=g=9.8 \mathrm{~m} / \mathrm{s}^2$
Final velocity of the ball $=\mathrm{v}$
From second equation of motion, time $(t)$ taken by the ball to hit the ground can be obtained as:
$\mathrm{s}=\mathrm{ut}+(1 / 2) \mathrm{at}^2$
$90=0+(1 / 2) \times 9.8 \mathrm{t}^2$
$\mathrm{t}=\sqrt{18.38}=4.29 \mathrm{~s}$
From first equation of motion, final velocity is given as:
$v=u+a t$
$=0+9.8 \times 4.29=42.04 \mathrm{~m} / \mathrm{s}$
Here we see that $\mathrm{v}=\mathrm{at}$, so velocity varies linearly with downward motion of ball.
This part of motion is represented by first line $(1)$ in the below graph
- First rebound
Now Rebound velocity of the ball, $u_r=v(1-1 / 10)=9 \mathrm{v} / 10=9 \times 42.04 / 10=37.84 \mathrm{~m} / \mathrm{s}$
This is represented the line $(2)$ in the graph. We are assumed negligible time of collision between ball and floor
- Upward motion of ball
Time $\left(t_1\right)$ taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
$\mathrm{v}=\mathrm{u}+\mathrm{at}_1$
$0=37.84+(-9.8) \mathrm{t}_1$
$\mathrm{t}_1=-37.84 /-9.8=3.86 \mathrm{~s}$
Total time taken by the ball $=\mathrm{t}+\mathrm{t}_1=4.29+3.86=8.15 \mathrm{~s}$
This is represented by line $(3)$ in the graph
- Second descent of ball
As the time of ascent is equal to the time of descent, the ball takes $3.86 s$ to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor $=9 \times 37.84 / 10=34.05 \mathrm{~m} / \mathrm{s}$
Total time taken by the ball for second rebound $=8.15+3.86=12.01 \mathrm{~s}$
This is represented by line $(4)$ in the graph View full question & answer→Question 41 Mark
Draw position-time graph for a body at rest.
Answer
In the above x-t graph, body is at rest at position $x_o$. View full question & answer→Question 51 Mark
Is it possible to have negative value in speed and displacement?
AnswerSpeed is always +ve. Displacement can have -ve, 0 or +ve values.
View full question & answer→Question 61 Mark
Why the speed of an object can never be negative?
AnswerSpeed is distance covered per unit time. Since distance cannot be negative therefore speed cannot be negative.
View full question & answer→Question 71 Mark
Write an expression for distance covered in $n^{th}$ second for a uniformly accelerated motion.
AnswerIf a is the uniform acceleration, then $\text{s(n}^{\text{th}})=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$ where, s ($n^{th}$) is the distance covered in $n^{th}$ second, u is the initial velocity.
View full question & answer→Question 81 Mark
The length covered by a body is found to be directly proportional to the square of time. What is the nature of acceleration?
Answer$\text{x}\propto\text{t}_2\therefore\text{v}\propto\text{t}$ and $\text{a}\propto\text{t}^0$ (i. e.) acceleration is constant.
View full question & answer→Question 91 Mark
Draw v-t graph for non-uniform accelerated motion.
Answer
Here, acceleration is increasing.
View full question & answer→Question 101 Mark
A car moving at a speed of 10m/s is accelerated at the rate of $2m/s^2$. Find out the velocity after 6 sec.
AnswerVelocity after 6 seconds $=10+2 \times 6=22 \mathrm{~ms}^{-1}$.
View full question & answer→Question 111 Mark
Two particles A and B start from rest and move for equal time on a straight line. The particle A has an acceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered larger distance?
AnswerSecond particle
Area under (v, t) graph gives displacement. For graph (2) area is more.
View full question & answer→Question 121 Mark
Speed of a particle cannot be negative. Why?
AnswerSpeed is the distance travelled in unit time and distance cannot be negative. negative.
View full question & answer→Question 131 Mark
Draw position-time graph for a non-uniform motion, when body starts from origin with increasing velocity.
Answer
In the above x-t graph, the slope of x-t graph is increasing, it means that the velocity is increasing.
View full question & answer→Question 141 Mark
Why does the earth impart the same acceleration to all bodies?
Question 151 Mark
It is a common observation that rain clouds can be at about a kilometre altitude above the ground. Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
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. Force exerted by a rain drop is $\text{F}=\frac{\text{Change in momentum}}{\text{Time}}=\frac{\text{p}-0}{\text{t}}$ $=\frac{4.7\times10^{-3}}{2.8\times10^{-5}}=168\text{N}$
View full question & answer→Question 161 Mark
Is it possible that a body could have constant speed but varying velocity?
AnswerYes, a body could have constant speed but varying velocity if only the direction of motion changes.
View full question & answer→Question 171 Mark
Which speed is measured by speedometer of your scooter?
AnswerInstantaneous speed of the scooter is measured by the speedometer.
View full question & answer→Question 181 Mark
What can you say about the nature of acceleration, associated with a mass whose y-t graph is shown?
AnswerSlope covers the same magnitude. So acceleration is constant.
View full question & answer→Question 191 Mark
Does the displacement of an object depend on the choice of the position of origin of the coordinate system?
AnswerNo, the displacement of the object does not depend on the choice of the position of the origin.
View full question & answer→Question 201 Mark
When a body accelerates by $\alpha \text{ t},$ what is the velocity after time 't', when it starts from rest?
Answer$\text{a}=\alpha\text{t}\text{ i.e.}, \int\text{dv}=\alpha\int\text{t dt}$$\nu=\frac{\alpha \text{ t}^2}{2}.$
View full question & answer→Question 211 Mark
Write the expression for distance covered in $n^{th}$ the second by a uniformly accelerated body.
AnswerIf a is the uniform acceleration then, $\text{s}=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$
View full question & answer→Question 221 Mark
Consider that the acceleration of a moving body varies with time. What does the area under acceleration-time graph for any time interval represent?
AnswerThe area under acceleration-time graph for any time interval represents the change of velocity of the body during that time interval.
View full question & answer→Question 231 Mark
What is the significance of the slope of x-t graph.
AnswerSlope of x-t graph provides velocity of motion. The nature of motion is identified by the shape of the graph.
View full question & answer→Question 241 Mark
Find the acceleration and velocity of a ball at the instant it reaches its highest point if it was thrown up with velocity v.
AnswerAcceleration is $9.8m/s^2$ (downwards) and velocity is zero at the highest point.
View full question & answer→Question 251 Mark
When a ball hits a wall with a velocity of 50m/s and bounces back with the same speed, what is the change in velocity of the ball?
Answer$\Delta \text{p}=-2\text{mv}=-100\text{m}$ $\therefore\text{m}\Delta\text{v}'=-100\text{m}$ $\Rightarrow \text{v}'=-100\text{m/s}$
View full question & answer→Question 261 Mark
What is the ratio of the time taken to go up and come down by a body thrown vertically up?
Answer1 : 1 in the absence of air resistance.
View full question & answer→Question 271 Mark
Two masses in the ratio 1 : 2 are thrown vertically up with the same speed. What is the effect on the time by the mass?
AnswerMass does not influence time.
View full question & answer→Question 281 Mark
What is the condition for an object to be considered as a point object?
AnswerAn object can be considered as a point object, if the distance travelled by it is very large than its size.
View full question & answer→Question 291 Mark
Can Earth be regarded as a 'point object' if only the orbital motion of Earth around the sun is considered?
AnswerYes. This is because the size of the Earth is very small as compared to the size of the orbit of the Earth around the sun.
View full question & answer→Question 301 Mark
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. At what time is its acceleration maximum in magnitude?
AnswerGiven velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ In a periodic motion when velocity is zero acceleration will be maximum putting v = 0 in Eq. (i). $0=6\text{t}-2\text{t}^2\Rightarrow0=\text{t}(6-2\text{t})$ $=\text{t}\times2(3-\text{t})=0\Rightarrow\text{t}=0\ \ \text{or}\ \ 3\text{s}$
View full question & answer→Question 311 Mark
Can a body have constant speed but a varying velocity?
AnswerYes, it is possible only if direction changes.
View full question & answer→Question 321 Mark
Do the following two graphs represent same type of motion? Name the motion.
AnswerBoth the graphs represent the non-uniform type of motion.
View full question & answer→Question 331 Mark
A ball is thrown straight up. What is its velocity and acceleration at the top?
Question 341 Mark
Give an example of a body possessing zero velocity and still accelerating.
AnswerA body thrown up vertically has zero velocity at the top-most point, but has acceleration of g.
View full question & answer→Question 351 Mark
The displacement-time graph for two particles X and Y are straight lines making angles of 30° and 60° with the time axis. What is the ratio of the velocities of Y and X?
Answer$\frac{\text{v}_\text{Y}}{\text{v}_\text{X}}=\frac{\tan60^\circ}{\tan30^\circ}$ $=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=3:1$
View full question & answer→Question 361 Mark
For which condition, the magnitude of average velocity is equal to the average speed for a particular motion?
AnswerWhen a particle is moving along a straight line with the fixed direction.
View full question & answer→Question 371 Mark
The position coordinate of a moving particle is given by $\mathrm{x}=6+18 \mathrm{t}+9 \mathrm{t}^2$, where x is in metres and t in seconds. What is the velocity at $\mathrm{t}=2 \mathrm{~s}$ ?
AnswerGiven, $\text{x}=6+18\text{t}+9\text{t}^2$ $\text{v}_\text{t}=\frac{\text{dx}}{\text{dt}}=18+18\text{t}$ At t = 2, $\text{v}_2=18+18\times2$ $=54\text{m/s}$
View full question & answer→Question 381 Mark
Two particles A and B are moving along the same straight line. Bis ahead of A. Velocities remaining unchanged, what would be effect on the magnitude of relative velocity if A ahead of B?
AnswerThere will be no effect on the magnitude of relative velocity.
View full question & answer→Question 391 Mark
At which point on its path a projectile has the smallest speed?
Question 401 Mark
What is the ratio of SI to CGS unit of acceleration?
AnswerRatio of SI to CGS unit of acceleration is $10^2$.
View full question & answer→Question 411 Mark
When a body accelerates by $\beta \text{t},$ what is the velocity after time t, when it starts from rest?
AnswerGiven, acceleration $\text{a}=\beta \text{}t$ It can be written as $\int\text{dv}=\int \beta\text{tdt}$ On intergrating, we get $\text{v}=\frac{\beta\text{t}^2}{2}+\text{C}=\frac{\beta\text{t}^2}{2}$ [$\because$ C = 0]
View full question & answer→Question 421 Mark
A mass is dropped from certain height. At the same time another equal mass is thrown with a horizontal velocity of u m/s. Which one of the two will reach the ground first?
AnswerBoth of them will reach the ground at the same time.
View full question & answer→Question 431 Mark
Shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
AnswerNo, because the x-t graph does not represent the trajectory of the path followed by a particles. From the graph, it is noted that at t = 0, x = 0.
View full question & answer→Question 441 Mark
The position coordinate of a moving particle is given by $x=6+18 t+9 t^2$ ( $x$ in metres and $t$ in seconds). What is its velocity at $\mathrm{t}=2 \mathrm{sec}$ ?
Answer$\mathrm{x}=6+18 \mathrm{t}+9 \mathrm{t}^2, \mathrm{v}=18+18 \mathrm{t} \mathrm{~v}_{\mathrm{t}}=2=18+18 \times 2=54 \mathrm{~ms}^{-1}$
View full question & answer→Question 451 Mark
Write the expression for distance covered in $n^{th}$ second by a uniformly accelerated body.
AnswerIf a is the uniform acceleration, then $\text{s}=\text{u}+\frac{\text{a}}{2}(2\text{n}-1),$ where u is the initial velocity.
View full question & answer→Question 461 Mark
The displacement-time graph of a particle is parallel to time-axis, what is the velocity of the particle?
AnswerThe velocity of the particle is zero because the slope of (x-t) graph is zero.
View full question & answer→Question 471 Mark
Is it possible that a body have a constant velocity but varying speed?
AnswerNo, because velocity is a speed with direction. Therefore, a body having constant velocity cannot have a varying speed.
View full question & answer→Question 481 Mark
What does speedometer record: the average speed or the instantaneous speed?
AnswerThe speedometer measures the instantaneous speed.
View full question & answer→Question 491 Mark
Which vector can be associated with a plane area? And what is its direction?
AnswerArea vector, outward drawn Normal to the plane will be its direction.
View full question & answer→Question 501 Mark
Write two uses of $v-t$ graph?
Answer
- Slope of $v-t$ graph gives acceleration.
- Area under $v-t$ graph gives displacement.
View full question & answer→Question 511 Mark
When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal. Comment on this statement.
AnswerIts average velocity and instantaneous velocity will be equal but we cannot compare these with average speed as one is vector and other is scalar.
View full question & answer→Question 521 Mark
A ball dropped from height h reaches the ground in t s. After what time the ball was passing through a point at a height $\frac{\text{h}}{2}$?
Answer$\text{t}=\sqrt{2\frac{\text{h}}{\text{g}}}$ $\text{t}'=\sqrt{\frac{1}{\text{g}}.\Big(\frac{\text{h}}{2}\Big)}=\sqrt{\frac{\text{h}}{\text{g}}}$ $\Rightarrow \text{t}'=\frac{\text{t}}{2}$
View full question & answer→Question 531 Mark
Suppose two trains A and B are moving with uniform velocities along parallel tracks in the same direction and the velocities of A and B be 60km/h in East and 65km/h in East. Find the relative velocity of B w.r.t. A.
AnswerRelative velocity of Bw.r.t. $A, V_{A B}=V_A-V_B$
View full question & answer→Question 541 Mark
The velocity of a particle is towards west at an instant. Its acceleration is not towards west, not towards east, not towards north and not towards south. Give an example of this type of motion.
AnswerProjectile at highest point because at the highest point vertical velocity become zero.
View full question & answer→Question 551 Mark
Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
AnswerThe given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
View full question & answer→Question 561 Mark
For which condition, the average velocity will be equal to the instantaneous velocity?
AnswerWhen a body moves with a uniform velocity, then $V_{av} = V_{ist}$
View full question & answer→Question 571 Mark
What is the shape of displacement-time graph for uniform linear motion?
AnswerA straight line inclined to the time axis.
View full question & answer→Question 581 Mark
Express an acceleration of $10m/ s^2$ in $km/ h^2$.
AnswerAcceleration$=\frac{10\text{m}}{(1\text{s)}^2}=\frac{10\times10^{-3}}{\Big[\frac{1}{60\times60}\text{h}\Big]^2}$ $=(3600)^2\times10^{-2}\text{km}/\text{ h}^2$ $=1.29\times10^5\text{km}/\text{ h}^2$
View full question & answer→Question 591 Mark
What will happen to a hydrogen balloon released on the moon?
AnswerThe balloon will fall with an acceleration of $g/6ms^{-2}$.
View full question & answer→Question 601 Mark
For which condition, the distance and the magnitude of displacement of an object have the same values?
AnswerThe distance and the magnitude of displacement of an object have the same values, when the body is moving along a straight line path in a fixed direction.
View full question & answer→Question 611 Mark
Why does a parachute descend slowly?
AnswerDue to increased surface area of a parachute, the air resistance is more, so it descends slowly.
View full question & answer→Question 621 Mark
If position of a particle at instant t is given by $x = 2t^3$, find the acceleration of the particle.
AnswerGiven, $x = 2t^3$, velocity, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}(2\text{t}^3)}{\text{dt}}=6\text{t}^2$ $\therefore$ Accceleration, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}(6\text{t}^2)}{\text{dt}}=12\text{t}$
View full question & answer→Question 631 Mark
Give an example of uniformly accelerated linear motion.
AnswerMotion of a body under gravity.
View full question & answer→Question 641 Mark
Draw velocity-time graph for an object, starting from rest. Acceleration is constant and remains positive.
Question 651 Mark
Can a moving body have relative velocity zero with respect to another body? Give an example.
AnswerYes, two cyclists moving with same velocity in the same direction.
View full question & answer→Question 661 Mark
The position x of a body is given by $\text{x}=\text{A}\sin(\omega\text{t}).$ Find the time at which the displacement is maximum.
AnswerThe value of position x will be maximum, when the value of sin (t) is maximum, for this $\sin(\omega\text{t})=1=\sin\frac{\pi}{2}$ $\omega\text{t}=\frac{\pi}{2}$ $\Rightarrow \text{t}=\Big(\frac{\pi}{2\omega}\Big)$
View full question & answer→Question 671 Mark
A man is standing on top of a building $100m$ high. He throws two balls vertically, one at $t = 0$ and other after a time interval $($less than $2$ seconds$)$. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15m$ at $t = 2s$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
AnswerWe solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take difference of displacements.
Let the speeds of the two balls $(1$ and $2)$ be $v_1$ and $v_2$ where:
if $\text{v}_1=2\text{v},\text{v}_2=\text{v}$ if $y_1$ and $y_2$ and the displacement covered by the balls $1$ and $2$, respectively,
before coming to rest, then, $\text{y}_1=\frac{\text{v}_1^2}{2\text{g}}=\frac{4\text{v}^2}{2\text{g}}$ and $\text{y}_2=\frac{\text{v}_2^2}{2\text{g}}=\frac{\text{v}^2}{2\text{g}}$
Since $\text{y}_1-\text{y}_2=15\text{m},\frac{4\text{v}^2}{2\text{g}}-\frac{\text{v}^2}{2\text{g}}=15\text{m or }\frac{3\text{v}^2}{2\text{g}}=15\text{m}$ or $\text{v}^2=\sqrt{5\text{m}\times(2\times10)}\text{m/s}^2$ or $\text{v}=10\text{m/s}$
Clearly, $v_1 = 20m/s$ and $v_2 = 10m/s$ as $\text{y}_1=\frac{\text{v}_1^2}{2\text{g}}=\frac{(20\text{m})^2}{2\times10\text{m}}=20\text{m}$ .
$\text{y}_2=\text{y}_1-15\text{m}=5\text{m}$ If $t_2$ is the time taken by the ball $2$ to cover a displacement of $5m$, then from
$\text{y}_2=\text{v}_2\text{t}-\frac{1}{2}\text{gt}^2_2$
$5=10\text{t}_2-5\text{t}^2_2\ \text{or}\ \text{t}^2_2-2\text{t}_2+1=0$
where $\text{t}_2=1\text{s}$ Since $t_1 ($time taken by ball $1$ to cover distance of $20m)$ is $2s$,
time interval between the two throws $=\text{t}_1-\text{t}_2=2\text{s}-1\text{s}=1\text{s}$
Important note: We should be very careful when we are applying the equation of rectilinear motion.
These equations are applicable only in case of constant acceleration. Some important observations for motion under gravity:
- The motion is independent of the mass of the body, as in any equation of motion, mass is not involved.
That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity, i.e., $\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}$ and $\text{v}=\sqrt{2\text{gh}}.$
- In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.
Time of descent $(t_1)$ = time of ascent $(\text{t}_2)=\frac{\text{u}}{\text{g}}$
Total time of flight $\text{T}=\text{t}_\text{x}+\text{t}_2=\frac{2\text{u}}{\text{g}}$
- In case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
As well as the magnitude of velocity at any point on the path is same whether the body is moving in upwards or downward direction.
- A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent is less than the time of descent $t_2> t_1$
Let u be the initial velocity of body, then time of ascent $\text{t}_1=\frac{\text{u}}{\text{g}+\text{a}}$ and $\text{h}=\frac{\text{u}^2}{2(\text{g}+\text{a})}$
where $g$ is acceleration due to gravity and $a$ is retardation by air resistance and for upward motion both will act vertically downward.
For downward ‘motion $a$ and $g$ will act in opposite direction because a always act in direction opposite to motion and $g$ always act vertically downward.
So, $\text{h}=\frac{1}{2}(\text{g}-\text{a})\text{t}^2_2$
$\Rightarrow\frac{\text{u}^2}{2(\text{g}+\text{a})}=\frac{1}{2}(\text{g}-\text{a})\text{t}^2_2$
$\Rightarrow\text{t}_2=\frac{\text{u}}{\sqrt{(\text{g}+\text{a})(\text{g}-\text{a})}}$
Comparing $t_1$ and $t_2$ we can say that $\text{t}_2 > \text{t}_1,$ since $(\text{g}+\text{a})>(\text{g}-\text{a})$ View full question & answer→Question 681 Mark
When a body accelerates by at, what is the velocity after time ‘t', when it starts from rest?
Answer$\text{a}=\alpha\text{t}\text{ i.e. },\int\text{dv}=\text{a}\int\text{t dt}$ $\Rightarrow \text{v}=\frac{\alpha\text{t}^2}{2}$
View full question & answer→Question 691 Mark
A truck and a car with the same kinetic energy are brought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance?
AnswerBoth truck and car will stop at the same distance.
View full question & answer→MCQ 701 Mark
The position of a particle moving in the $X-Y$ plane at any time $t$ is given by; $x=\left(3 t^2-6 t\right)$ meters; $y=\left(t^2-2 t\right)$ meters. Select the correct statement:
- A
Acceleration is zero at $t = 0$
- B
Velocity is zero at $t = 0$
- ✓
Velocity is zero at $t = 1$ second
- D
Velocity and acceleration of the particle are never zero
AnswerCorrect option: C. Velocity is zero at $t = 1$ second
$\text{x}=3\text{t}^2-6\text{t},$ So $($Velocity$)_x=\frac{\text{dx}}{\text{dt}}=6\text{t}-6$
$($Acceleration$)_x=\frac{\text{d}^2\text{y}}{\text{dt}^2}=6\text{t};\text{y}=\text{t}^2-2\text{t};$
so $($Velocity$)_y=\frac{\text{d}^2\text{y}}{\text{dt}^2}=2;$ At time $t = 1,$
$\frac{\text{dx}}{\text{dt}}=6\times1-6=0$ and $\frac{\text{dy}}{\text{dt}}=2\times1-2=0$
View full question & answer→Question 711 Mark
Define one dimensional motion.
AnswerA particle moving along a straight line or a path is said to undergo one dimensional motion.
View full question & answer→Question 721 Mark
Why is it not necessary for a body following another to stop, to avoid collision?
AnswerIf the relative velocity becomes zero, the collision can be avoided.
View full question & answer→Question 731 Mark
Displacement-time (S-T) graph of any object is shown in figure. Draw velocity-time graph for this motion.
Question 741 Mark
A car starts accelerating from rest for sometime, maintains the velocity for sometime and then comes to rest with uniform deceleration. Draw V-t graph.
Question 751 Mark
State the condition when the magnitude of velocity and speed of an object are equal.
AnswerWhen a body moves in a straight line.
View full question & answer→Question 761 Mark
What does the slope of velocity-time graph represent?
AnswerThe slope of velocity-time graph represents acceleration.
View full question & answer→Question 771 Mark
Define displacement of a particle.
AnswerThe change in the position co-ordinates of a particle over a given period of time is called the displacement of the particle.
View full question & answer→Question 781 Mark
Define uniformly accelerated motion.
AnswerA body has uniformly accelerated motion if it moves with constant acceleration.
View full question & answer→Question 791 Mark
A ball is dropped from a height of $90m$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed$-$time graph of its motion between $t = 0$ to $12s$.
Answer
- Downward motion of Ball
Ball is dropped from a height, $s=90 \mathrm{~m}$
Initial velocity of the ball, $\mathrm{u}=0$
Acceleration, $\mathrm{a}=\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
Final velocity of the ball $=\mathrm{v}$
From second equation of motion, time $(t)$ taken by the ball to hit the ground can be obtained as:
$s=u t+(1 / 2) \mathrm{at}^2$
$90=0+(1 / 2) \times 9.8 \mathrm{t}^2$
$\mathrm{t}=\sqrt{18.38}=4.29 \mathrm{~s}$
From first equation of motion, final velocity is given as:
$v=u+a t$
$=0+9.8 \times 4.29=42.04 \mathrm{~m} / \mathrm{s}$
Here we see that $v=a t$, so velocity varies linearly with downward motion of ball.
This part of motion is represented by first line $(1)$ in the below graph
- First rebound
Now Rebound velocity of the ball, $\mathrm{u}_{\mathrm{r}}=\mathrm{v}(1-1 / 10)=9 \mathrm{v} / 10=9 \times 42.04 / 10=37.84 \mathrm{~m} / \mathrm{s}$
This is represented the line $(2)$ in the graph. We are assumed negligible time of collision between ball and floor
- Upward motion of ball
Time $\left(t_1\right)$ taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
$\mathrm{v}=\mathrm{u}+\mathrm{at}_1$
$0=37.84+(-9.8) \mathrm{t}_1$
$\mathrm{t}_1=-37.84 /-9.8=3.86 \mathrm{~s}$
Total time taken by the ball $=\mathrm{t}+\mathrm{t}_1=4.29+3.86=8.15 \mathrm{~s}$
This is represented by line $(3)$ in the graph
- Second descent of ball
As the time of ascent is equal to the time of descent, the ball takes $3.86 s$ to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor $=9 \times 37.84 / 10=34.05 \mathrm{~m} / \mathrm{s}$
Total time taken by the ball for second rebound $=8.15+3.86=12.01 \mathrm{~s}$
This is represented by line $(4)$ in the graph View full question & answer→Question 801 Mark
A car travels at a speed of 60km/hr due north and the other at a speed of 60km/hr due east. Are the velocities equal? If no, which one is greater? If you find any of the questions irrelevant, explain.
AnswerThere are equal because they have different directions. There magnitude can be compared.
View full question & answer→Question 811 Mark
If in case of a motion, displacement is directly proportional to the square of time elapsed, what do you think about its acceleration i.e., constant or variable? Explain why?
AnswerAcceleration is constant, since the equation of motion canbe applied only then.
View full question & answer→Question 821 Mark
Constant acceleration means that x-t graph will have constant slope? Yes/ No
AnswerAcceleration means that velocity is non-uniform. So, x-t graph will be curved.
View full question & answer→Question 831 Mark
Define non-uniformly accelerated motion.
AnswerA body has non-uniformly accelerated motion if it moves with variable acceleration.
View full question & answer→Question 841 Mark
Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.
You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 2.8). After you catch it, find the distance $d$ travelled by the ruler. In a particular case, $d$ was found to be $21.0 cm$. Estimate reaction time.
AnswerThe ruler drops under free fall. Therefore, $v_o=0$, and $a=-g=-9.8 m s ^{-2}$.
The distance travelled $d$ and the reaction time $t_r$ are related by
$d=-\frac{1}{2} g t_r^2$
$t_r=\sqrt{\frac{2 d}{g}} s$
Given $d=21.0 \ cm$ and $g=9.8 m s ^{-2}$ the reaction time is
$t_r=\sqrt{\frac{2 \times 0.21}{9.8}} s \equiv 0.2 s .$
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