Question 12 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A spinning cricket ball that turns sharply on hitting the ground.
AnswerThe size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
View full question & answer→Question 22 Marks
A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$. What is the direction of acceleration during the upward motion of the ball?
AnswerIrrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
View full question & answer→Question 32 Marks
Read statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion. With zero speed may have non-zero velocity.
AnswerFalse. Explanation: Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.
View full question & answer→Question 42 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A tumbling beaker that has slipped off the edge of a table.
AnswerThe size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
View full question & answer→Question 52 Marks
Suggest a suitable physical situation for of the following graphs.
AnswerIn the given v-tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
View full question & answer→Question 62 Marks
Suggest a suitable physical situation for of the following graphs.
AnswerThe given a-t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.
View full question & answer→Question 72 Marks
A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$. What are the velocity and acceleration of the ball at the highest point of its motion?
AnswerAt maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 \mathrm{~m} / \mathrm{s}^2$.
View full question & answer→Question 82 Marks
Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
AnswerSpeed has negative values in the graph which is not possible. So, the given v-t graph, shown in (c), does not represent one-dimensional motion of the particle.
View full question & answer→Question 92 Marks
Read statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion. With zero speed at an instant may have non-zero acceleration at that instant.
AnswerTrue. Explanation: When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
View full question & answer→Question 102 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A monkey sitting on top of a man cycling smoothly on a circular track.
AnswerThe size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
View full question & answer→Question 112 Marks
A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 \mathrm{~km}{ }^{\mathrm{h}-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 \mathrm{~km} \mathrm{~h}^{-1}$. What is the Magnitude of average velocity, and.
AnswerTime taken by the man to reach the market from home, $\mathrm{t}_1=2.5 / 5=1 / 2 \mathrm{~h}=30 \mathrm{~min}$ Time taken by the man to reach home from the market, $\mathrm{t}_2=2.5 / 7.5=1 / 3 \mathrm{~h}=20 \mathrm{~min}$ Total time taken in the whole journey $=30+20=50 \mathrm{~min}$
View full question & answer→Question 122 Marks
Read statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion. With constant speed must have zero acceleration.
AnswerTrue. Explanation: A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.
View full question & answer→Question 132 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A railway carriage moving without jerks between two stations.
AnswerThe size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
View full question & answer→Question 142 Marks
Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
AnswerIf a vertical line is drawn from any point on axis representing t, we see that it cuts the graph at two points which means particle has two velocities at the same instant of time which is not possible. So, the given v-t graph, shown in (b), does not represent one-dimensional motion of the particle.
View full question & answer→Question 152 Marks
Look at the graphs carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
AnswerIf a vertical line is drawn from any point on axis representing t, we see that it cuts the graph at two points which means particle has two positions at the same instant of time which is not possible. So, the given x-t graph, shown in (a), does not represent one-dimensional motion of the particle.
View full question & answer→Question 162 Marks
Under what condition will the distance and displacement of a moving object will have the same magnitude?
AnswerDistance and displacement will have the same magnitude when the object moves along a straight line without change in its direction.
View full question & answer→Question 172 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A spinning cricket ball that turns sharply on hitting the ground.
AnswerThe size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
View full question & answer→Question 182 Marks
A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$. What is the direction of acceleration during the upward motion of the ball?
AnswerIrrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
View full question & answer→Question 192 Marks
Prove that $\vec{\text{v}}_\text{AB}=\vec{\text{v}}_\text{A}-\vec{\text{v}}_\text{B},$ where symbols have their usual meaning.
AnswerConsider two bodies A and B moving with velocities VA and vg. Let x ; and xz ; be their initial positions. Their position after time $t$ is given by, $x_{1 f}=x_{1 i}+v_{A t} x_{2 f}=x_{2 i}+v_B t$ Subtacting, $\left(x_{1 f}-x_{2 f}\right)=\left(x_{1 i}-x_{2 i}\right)+t\left(v_A-v_B\right)\left(x_{1 f}-x_{2 f}\right)-\left(x_{1 i}-x_{2 i}\right)=$
$t\left(v_A-v_B\right)$ The change in separation in time ' $t$ ' is $t\left(v_A-v_B\right) \cdot\left(v_A-v_B\right)$ is called relative velocity of $A$ with respect to $B$. It is nothig but the change in separation per unit time.
View full question & answer→Question 202 Marks
The acceleration-time graph for a body is shown in the figure below. Plot the corresponding velocity-time graph.
AnswerFor time t, there is an acceleration a. $\therefore$ Velocity increases after time 't' acceleration is zero. Thus v-t graph will be shown as 
View full question & answer→Question 212 Marks
From the given example, find if the motion is one or two or three$-$dimensional.
- A kite flying in the sky.
- A cricket ball hit by a player.
- Moon revolving around the earth.
- The motion of a stone in a circle.
Answer
- A flying kite in the sky comes under three$-$dimensional motion.
- A cricket ball hit by a player comes under two$-$dimensional motion.
- Moon revolving around the sun$-$earth comes under two$-$dimensional motion.
- The motion of the stone in circular motion comes under two$-$dimensional motion.
View full question & answer→Question 222 Marks
For what condition, an object could be considered as a point object? Describe in brief.
AnswerAn object could be considered as a point object if it covers a distance much larger than its own size. e.g. If a bus of 5m in size move 100km, then the bus can be considered as a point object.
View full question & answer→Question 232 Marks
The position of an object is given by $x=2 t^2+3 t$. Find out that its motion is uniform and non-uniform.
AnswerAs given, $x=2 t^2+3 t$ By differentiating x w.r.t. t, we get Velocity, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(2\text{t}^2+3\text{t})$ $\text{v}=(4\text{t}+3)$ As velocity is time dependent, it means that motion is non-uniform.
View full question & answer→Question 242 Marks
Two particles begin to fall freely from rest from the top of a tower within a gap of 1 s. How long after the first particle begins to fall, the two particles be 15m apart? (Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
AnswerLet the first particle takes time to reach a position co-ordinate 15m below the second particle. Obviously, the second body has fallen under gravity for a time (t-1) s only. Hence $15=\text{y}_1-\text{y}_2=\frac{1}{2}\text{gt}^2-\frac{1}{2}\text{g}(\text{t}-1)^2=\text{gt}-\frac{\text{g}}{2}$ $15=10\text{t}-5$ $\Rightarrow\text{t}=\frac{15+15}{10}=2\text{s}$
View full question & answer→Question 252 Marks
The two straight rays $OA$ and $OB$ on the same displacement $-$ time graph make angle $30^\circ$ and $60^\circ$ with time axis respectively as shown in figure.
- Which ray represents greater velocity?
- What is the ratio of two velocities represented by $OA$ and $OB?$
Answer
- $OB$, because the slope has greater angle.
- Ratio of two velocities
$\frac{\text{v}_\text{A}}{\text{v}_\text{B}}=\frac{\tan30^\circ}{\tan^\circ}$
$=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}$ View full question & answer→Question 262 Marks
Give examples where:
- The velocity of a particle is zero but its acceleration is not zero.
- The velocity is opposite in direction to the acceleration.
- The velocity is perpendicular to the acceleration.
AnswerParticle thrown upwards:
- At highest point.
- While going up.
- At the highest point of projectile.
View full question & answer→Question 272 Marks
A body starts from rest and travels in straight line with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$ for 5 seconds. What is the velocity and distance travelled in this time?
AnswerVelocity after 5 seconds $=\mathrm{v}=\mathrm{u}+\mathrm{at}=0+5 \times 5=25 \mathrm{~ms}^{-1}$ Distance travelled in 5 seconds $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$ $=0+\frac{1}{2} \times 5 \times 5^2=62.5 \mathrm{~m}$.
View full question & answer→Question 282 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A tumbling beaker that has slipped off the edge of a table.
AnswerThe size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
View full question & answer→Question 292 Marks
What is common between the two graphs shown in Figs. (a) and (b)?
AnswerBoth represent negative velocity.
View full question & answer→Question 302 Marks
A ball is thrown vertically upward with a speed of 20m/s. Draw a graph showing the velocity of the ball as a function of time as it goes up and then comes back.
Question 312 Marks
Find the area enclosed by the curve $\text{y}=\sin\text{x}$ and the X-axis between $\text{x}=0$ and $\text{x}=\pi.$
Answer$\text{Area}=\int\limits^{\text{y}}_{0}\text{dy}=\int\limits^{\pi}_0\sin\text{x dx}=-[\cos\text{x}]^\pi_0=2$
View full question & answer→Question 322 Marks
Why does a person sitting in one train think that the other train is at rest, when both trains are moving on parallel tracks with the same speed and in the same direction?
AnswerThis is because the relative velocity of the train in which the person is sitting w.r.t. the other train is zero. For example, let the velocity of A train = 60km/ hr due east Velocity of B train = 60km/ hr due east $\therefore$ Relative velocity of A train w.r.t. B train = 60 - 60 = 0
View full question & answer→Question 332 Marks
For the motion shown in the figure, find the displacement of car between the time intervals $t_1$ and $t_3$. 
AnswerCar moving towards +x-axis. At time $t_1$ it is at +20m. It turns at +50m and starts moving towards - x-axis. At time $t_3$, it reached at -30m. The displacement of car in the interval $(\text{t}_3-\text{t}_1)=\Delta\text{t}$ is -30-(+20) = -50m (negative sign shows the direction of displacement is towards - x-axis).
View full question & answer→Question 342 Marks
An object moving on a straight line covers first half of the distance at speed $v$ and second half of the distance at speed $2v.$ Find:
- Aaverage speed.
- Mean speed.
AnswerLet total distance be $x.$
Distance of first half $=\frac{\text{x}}{2},$ Speed $= v$
Time taken $t_1=\frac{\frac{\text{x}}{2}}{\text{v}}=\frac{\text{x}}{2\text{v}}$
Distance of second half $=\frac{\text{x}}{2},$
Speed $= 2v$
Time taken $t_2$ $=\frac{\frac{\text{x}}{2}}{2\text{v}}=\frac{\text{x}}{4\text{v}}$
- Average speed $=\frac{\text{Total distance travelled}}{\text{Total time taken}}$
$=\frac{\text{x}}{\frac{\text{x}}{2\text{v}}+\frac{\text{x}}{4\text{v}}}=\frac{4\text{v}}{3}$
- Mean speed $=\frac{\text{v}+2\text{v}}{2}=\frac{3\text{v}}{2}$
View full question & answer→Question 352 Marks
Is the speed-time graph shown in Fig possible?
AnswerNo. Speed cannot be negative.
View full question & answer→Question 362 Marks
A curve is represented by $\text{y}=\sin\text{x}.$ If x is changed from $\frac{\pi}{3}$ to $\frac{\pi}{3}+\frac{\pi}{100},$ find approximately the change in y.
Answer$\text{y}=\sin\text{x}$ So, $\text{y}+\triangle\text{y}=\sin(\text{x}+\triangle\text{x})$ $\triangle\text{y}=\sin(\text{x}+\triangle\text{x})-\sin\text{x}$ $=\Big(\frac{\pi}{3}+\frac{\pi}{100}\Big)-\sin\frac{\pi}{3}=0.0157$ 
View full question & answer→Question 372 Marks
Draw displacement-time graph for a uniformly accelerated motion. What is its shape?
AnswerDisplacement-time graph for a uniformly accelerated motion has been shown in adjoining Fig. The graph is parabolic in shape.
View full question & answer→Question 382 Marks
Can a particle have acceleration at an instant if its velocity is zero at that instant? Give example.
AnswerYes, it is possible. In motion under gravity at the highest point of its motion velocity is zero but acceleration a = g in downward direction.
View full question & answer→Question 392 Marks
The speed$-$time graph for a car is shown in figure below.
- Find how far the car travels in the first $4s?$ Shade the area on the graph that represents the distance travelled by the car during the period.
- Which part of the graph represent uniform motion of the car?
Answer
- The shaded portion of the car represents the distance travelled by the car in the first four seconds.
The car travels with a non$-$uniform speed which is accelerated in nature.
- The straight line portion of the graph represents the uniform motion of the car i.e. from point $A$ to $B.$
View full question & answer→Question 402 Marks
The changes in a function y and the independent dy variable x are related as $\frac{\text{dy}}{\text{dx}}=\text{x}^2.$ Find y as a function of x.
AnswerThe change in a function of y and the independent variable x are related as $\frac{\text{dy}}{\text{dx}}=\text{x}^2.$ $\Rightarrow\text{dy}=\text{x}^2\text{dx}$ Taking integration of both sides, $\int\text{dy}=\int\text{x}^2\text{dx}\Rightarrow\text{y}=\frac{\text{x}^3}{3}+\text{c}$ $\therefore$ y as a function of x is represented by $\text{y}=\frac{\text{x}^3}{3}+\text{c}.$
View full question & answer→Question 412 Marks
Can the relative velocity of two bodies be greater than the absolute velocity of either?
AnswerYes. For example, when two bodies move in opposite direction then relative velocity of each is greater than the individual velocities.
View full question & answer→Question 422 Marks
The speed time graph for a car is shown in Fig.
- Find how far does the car travel in the first $4$ seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
- Which part of the graph represents uniform motion of the car?
Answer
- Shaded part represents distance travelled in $4$ seconds.
Area of shaded part $=\frac{6\text{ms}^{-1}\times4\text{s}}{2}=12\text{m}$
- Straight part $AB$ represents a uniform motion of the car.
View full question & answer→Question 432 Marks
The displacement of a particle is given by $at^2$. What is dependency of acceleration on time?
AnswerLet x be the displacement. Then, $x = at^2 \therefore$ Velocity of the object, $\text{v}=\frac{\text{dx}}{\text{dt}}=2\text{at}$ Acceleration of the object, $\text{a}=\frac{\text{dv}}{\text{dt}}=2\text{a}$ It means that a is constant.
View full question & answer→Question 442 Marks
A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$. What are the velocity and acceleration of the ball at the highest point of its motion?
AnswerAt maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 \mathrm{~m} / \mathrm{s}^2$.
View full question & answer→Question 452 Marks
If the displacement of a body is zero, is distance necessarily zero? Answer with one example.
AnswerNo, because the distance covered by an object is the path length of the path covered by the object. The displacement of an object is given by the change in position between the initial position and final position. e.g. A boy starts from his home and moves towards market along a straight path. Then, he returns to home from the same path. Here, displacement is zero but distance is non-zero.
View full question & answer→Question 462 Marks
A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18m/h while the other has the speed of 27km/h. The bird starts moving from first car towards the other and is moving with the speed of 36km/h and when the two cars were separted by 36km. What is the total distance covered by the bird? What is the total displacement of the bird?
AnswerRelative speed of cars = 27 + 18 = 45Km/hr Time to meet the two cars together (t) $\text{t}=\frac{\text{distance between cars}}{\text{relative speed of cars}}=\frac{36\text{km}}{(27+18)\text{km/hr}}$ $=\frac{36}{45}=\frac{4}{5}\Rightarrow\text{t}=\frac{4}{5}\text{hours}$ $\therefore$ Distance covered by the bird in $\frac{4}{5}\text{hours}=36\times\frac{4}{5}=28.8\text{km.}$
View full question & answer→Question 472 Marks
Draw a graph from the following data. Draw tangents at $x = 2, 4, 6$ and $8$. Find the slopes of these tangents. Verify that the curve drawn is $y = 2x^2$ and the slope of dy tangent is $\tan\theta=\frac{\text{dy}}{\text{dx}}=4\text{x}.$
| x |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
| y |
2 |
8 |
18 |
32 |
50 |
72 |
98 |
128 |
162 |
200 |
AnswerThe graph $y = 2x^2$ should be drawn by the student on a graph paper for exact results. To find slope at any point, draw a tangent at the point and extend the line to meet x-axis. Then find $\tan\theta$ as shown in the figure. It can be checked that, Slope $=\tan\theta=\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(2\text{x}^2)=4\text{x}$ Where x = the x-coordinate of the point where the slope is to be measured.
View full question & answer→Question 482 Marks
A bus starting from rest moves with a uniform acceleration of $0.1m/s^2$ for $2$ min. Find:
- The speed acquired.
- The distance travelled.
Answer$u= 0, a = 0.1\ m/s$ and $t = 2$ min $120s$
- $v = u + at$
$= 0 + 0.1 \times 120$
$= 12m/s$
- $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
$=0+\frac{1}{2}\times0.1\times(120)^2$
$=\frac{1}{2}\times0.1\times120\times120$
$=720\text{m}$ View full question & answer→Question 492 Marks
The electric current in a charging $R-C$ circuit is given by
$\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
where $\text{i}_0, R$ and $C$ are constant parameters of the circuit and $t$ is time.
Find the rate of change of current at
- $t = 0.$
- $t = RC.$
- $t = 10RC.$
AnswerGiven that, $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
$\therefore$ Rate of change of current $=\frac{\text{di}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\text{i}_0\text{e}^{-\frac{\text{i}}{\text{RC}}}=\text{i}_0\frac{\text{d}}{\text{dt}}\text{e}^{-\frac{\text{t}}{\text{RC}}}=\frac{-\text{i}_0}{\text{RC}}\times\text{e}^{-\frac{\text{t}}{\text{RC}}}$
- When $\text{t}=0,\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RC}}$
- when $\text{t}=\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}}$
- when $\text{t}=10\text{RC},\frac{\text{di}}{\text{dt}}=\frac{-\text{i}}{\text{RCe}^{10}}$
View full question & answer→Question 502 Marks
A cyclist covers first half of a length with a speed of 5m/s and the second half with a speed of 10m/s. What is the average speed of the cyclist?
AnswerAverage Speed $=\frac{2\text{S}}{\frac{\text{S}}{5}+\frac{\text{S}}{10}}=\frac{2\times5\times10}{15}$ $=\frac{100}{15}\text{ms}^{-1}$
View full question & answer→Question 512 Marks
The distance travelled by a body is proportional to the square of time. What type of motion this body has?
AnswerHere, $\text{x}^2\propto \text{t}^2$ or $\text{x}=\text{kt}^2$ where , k is constant of proportinality. Now, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{kt}^2)=2\text{kt}$ and, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{}\text{dt}(2\text{kt})=2\text{k}$ (constant) Thus, the body has uniform accelerated motion.
View full question & answer→Question 522 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A monkey sitting on top of a man cycling smoothly on a circular track.
AnswerThe size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
View full question & answer→Question 532 Marks
A body goes from A to B with a velocity of 40m/s and comes back from B to A with a velocity of 60m/s. What is the (i) average velocity during the whole journey and (ii) average speed during the whole journey?
AnswerAverage velocity $=\frac{\text{Total displacement}}{\text{times}}=0$ Since, there is no net displacement Average speed $=\frac{\text{Total distance}}{\text{time}}$ $=\frac{2\text{AB}}{\frac{\text{AB}}{40}+\frac{\text{AB}}{60}}=48\text{m/s}$
View full question & answer→Question 542 Marks
It is a common observation that rain clouds can be at about a kilometre altitude above the ground. Estimate the time required to flatten the drop.
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. Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground. $\text{t}=\frac{\text{d}}{\text{v}}=\frac{4\times10^{-3}}{100\sqrt{2}}=0.028\times10^{-3}\text{s}$ $=2.8\times10^{-5}\text{s}=30\text{ms}$
View full question & answer→Question 552 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. At what time is its average velocity maximum?
AnswerGiven velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ From Eq. (i) $\text{v}=6\text{t}-2\text{t}^2$ $\Rightarrow\frac{\text{ds}}{\text{dt}}=6\text{t}-2\text{t}^2$ $\Rightarrow\text{ds}=(6\text{t}-2\text{t}^2)\text{dt}$ where, s is displacement $\therefore$ Distance travelled in time interval 0 to 3s, $\text{s}=\int^3_0(6\text{t}-2\text{t}^2)\text{dt}$ $=\Big[\frac{6\text{t}^2}{2}-\frac{2\text{t}^3}{3}\Big]^3_0=\Big[3\text{t}^2-\frac{2}{3}\text{t}^3\Big]^3_0$ $=3\times9-\frac{2}{3}\times3\times3\times3$ $=27-18=9\text{m}$ $\text{Average velocity}=\frac{\text{Distance travelled}}{\text{Time}}$ $=\frac{9}{3}=3\text{m/s}$ Given, $\text{x}=6\text{t}-2\text{t}^2$ $\Rightarrow3=6\text{t}-2\text{t}^2\Rightarrow2\text{t}^2-6\text{t}-3=0$ $\Rightarrow\text{t}=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}$ $=\frac{6\pm\sqrt{12}}{4}=\frac{3\pm2\sqrt{3}}{2}$ Considering positive sign only $\text{t}=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times1.732}{2}=\frac{9}{4}\text{s}$
View full question & answer→Question 562 Marks
What is common between two graphs shown below?
AnswerBoth graphs (a) and (b) represent positive and constant (different) velocity.
View full question & answer→Question 572 Marks
Acceleration is called as rate of change of velocity. Suppose we call rate of change of acceleration SLAP, what is the unit of SLAP?
AnswerSLAP $=\frac{\text{Acceleration}}{\text{Time}}$ $=\frac{\text{ms}^{-2}}{\text{s}}=\text{ms}^{-3}$
View full question & answer→Question 582 Marks
A body travels with a velocity $\mathrm{v}_1$ for time $\mathrm{t}_1$ and with a velocity $\mathrm{v}_2$ for time $\mathrm{t}_2$, find the average velocity of the body for the total time.
AnswerDisplacement travelled in time $(\text{t}_1+\text{t}_2)=\text{s}_1+\text{s}_2$ $=\text{v}_1\text{t}_1+\text{v}_2\text{t}_2$ $\therefore$ Average velocity $=\frac{\text{Net displacement}}{\text{Total time taken}}$ $=\frac{\text{v}_1\text{t}_1+\text{v}_2\text{t}_2}{(\text{t}_1+\text{t}_2)}$
View full question & answer→Question 592 Marks
The electric current in a discharging $R-C$ circuit is given by $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$ where $i_0, R$ and $C$ are constant parameters and $t$ is time. Let $\text{i}_0=2.00\text{A},\text{R}=600\times10^5\Omega$ and $\text{C}=0.500\mu\text{F}.$
- Find the current at $t = 0.3s.$
- Find the rate of change of current at $t = 0.3s.$
- Find approximately the current at $t = 0.31s.$
AnswerEquation $\text{i = i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
$\text{i}_0=2\text{A, R}=6\times10^{-5}\Omega,\text{C}=0.0500\times10^{-6},\text{F}=5\times10^{-7}\text{F}$
- $\text{i}=2\times\text{e}^{\Big(\frac{-0.3}{6\times0^3\times5\times10^{-7}}\Big)}=2\times\text{e}^{\big(\frac{-0.3}{0.3}\big)}=\frac{2}{\text{e}}\text{amp}$
- $\frac{\text{di}}{\text{dt}}=\frac{-\text{i}_0}{\text{RC}}\text{e}^{-\frac{\text{t}}{\text{RC}}}$ when $t = 0.3$
$\sec\Rightarrow\frac{\text{di}}{\text{dt}}=-\frac{2}{0.30}\text{e}^{\big(\frac{-03}{0.3}\big)}=\frac{-20}{3\text{e}}\text{amp/sec}$
- At $\text{t}=0.31\text{ sec},\text{i}=2\text{e}^{\big(-\frac{-0.3}{0.3}\big)}=\frac{5.8}{3\text{e}}\text{amp}.$
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A car travelling with a speed of 90km/ h on a straight road is ahead of a scooter travelling with a speed of 60km/ h. How would the relative velocity be altered, if scooter is ahead of the car?
AnswerLet $\mathrm{v}_{\mathrm{c}}$ and $\mathrm{v}_{\mathrm{s}}$ be the velocities of the car and the scooter, respectively. $\mathrm{v}_{\mathrm{C}}=90 \mathrm{~km} / \mathrm{h}$ and $\mathrm{v}_{\mathrm{s}}=60 \mathrm{~km} / \mathrm{h}$ (given) When the car is ahead of the scooter, then the relative velocity is $\mathrm{V}_{\mathrm{CS}}=\mathrm{v}_{\mathrm{C}}-\mathrm{v}_{\mathrm{s}}=90-60=30 \mathrm{~km} / \mathrm{h}$ (away from the scooter) When the scooter is ahead of the car, then the relative velocity is $\mathrm{V}_{\mathrm{Cs}}=\mathrm{v} .-\mathrm{v}_r=90-60=30 \mathrm{~km} / \mathrm{h}$ (towards the scooter)
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A car travelling at a speed of 10m/s due North, turns to its left and travels with same speed. Find the change in velocity associated.
AnswerChange in velocity, $\text{v}_\text{f}-\text{v}_\text{i}=\vec{\text{v}}_\text{W}-\vec{\text{v}}_\text{N}=\vec{\text{v}}_\text{W}\\+(-\vec{\text{v}}_\text{N})=\vec{\text{v}}_\text{W}+\vec{\text{v}}_{\text{S}}$ Since magnitudes are equal the change in velocity $=\sqrt{2}\text{v}_\text{W}=10\sqrt{2}$ in south westerly direction.
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The graph between total path length and time for a particle moving along a straight line as shown in figure is not possible. Explain why?
AnswerThe graph shows that with the passage of time, total path length first increases and then decreases. The path length always increases or remains constant with passage of time and it does not decrease with time as shown in figure. Thus, this graph is not possible.
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Which of the following is true for displacement?
- It cannot be zero.
- Its magnitude is greater than the distance travelled by the object.
AnswerBoth these statements are not true, because:
- Its magnitude can be zero.
- Its magnitude is either less than or equal to the distance travelled by the object.
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If a particle is accelerating, it is either speeding up or speeding down. Do you agree with this statement?
AnswerAcceleration doesnt mean speeding up or down. It means change of velocity either change in magnitude or direction.
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Find the area bounded under the curve $y = 3x^2 + 6x + 7$ and the X-axis with the ordinates at $x = 5$ and $x = 10$.
Answer$y = 3x^2 + 6x + 7 \therefore$ Area bounded by the curve, x-axis with coordinates with $x = 5$ and $x = 10$ is given by, $\text{Area}=\int\limits^{\text{y}}_{0}\text{dy}=\int\limits^{10}_5(3\text{x}^2+6\text{x}+7)\text{dx}\\=3\frac{\text{x}^3}{3}\Big]^{10}_5+5\frac{\text{x}^2}{3}\Big]^{10}_5+7\text{x}]^{10}_{5}=1135\text{sq. units}$
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An object is moving with uniform velocity $v$ along a straight line. What will be the shape of displacement$-$time graph if:
- $x_0 = -ve, v = +ve$,
- $x_0 = +ve, v = -ve$
Here, $x_0$ represents the position of the particle at time $t = 0$. AnswerDisplacement$-$time graphs are shown in Fig.
In Fig. $(a)\ x_0$ is $-ve$ but velocity $v$ is $+ve$
i.e., slope of curve is $+ve.$
In Fig. $(b), x_0$ is $+ve$ but slope of line and hence velocity is negative.
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A body is travelling in a straight line with a uniformly increasing speed. Plot a graph which represents the change in distance (s) travelled with time (t).
AnswerLet a body is travelling with initial speed ‘u’ in a straight line, then using relation, 
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$ For u = 0; $\text{s}=\frac{1}{2}\text{at}^2$ $\text{s}\propto \text{t}^2$ View full question & answer→Question 682 Marks
What is meant by 'point object' in physics?
AnswerAn object is said to be point object if its dimensions are negligible as compared to the distance travelled by it. For example, an aeroplane which flies from Delhi to London.
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A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 \mathrm{~km}{ }^{\mathrm{h}-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 \mathrm{~km} \mathrm{~h}^{-1}$. What is the Magnitude of average velocity, and.
AnswerTime taken by the man to reach the market from home $\mathrm{t}_1=2.5 / 5=1 / 2 \mathrm{~h}=30 \mathrm{~min}$ Time taken by the man to reach home from the market, $\mathrm{t}_2=2.5 / 7.5=1 / 3 \mathrm{~h}=20 \mathrm{~min}$ Total time taken in the whole journey $=30+20=50 \mathrm{~min}$
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A particle is moving in a straight line. Is it possible for it to maintain the motion in the same direction while the acceleration is in the reverse direction?
AnswerYes, due to acceleration in reverse direction the velocity starts decreasing with time but the direction of motion is maintained till the velocity is reduced to zero.
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A ball is thrown vertically upwards. Draw its:
- Velocity$-$time curve.
- Acceleration$-$time curve.
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At t = 0, a particle is at rest at origin. Its acceleration is $2m/s^2$ for the first 3s and $-2m/s^2$ for next 3s. Plot the acceleration versus time and velocity versus time graph.
AnswerThe acceleration-time graph is
The area enclosed between a-t curve gives change in velocity for the corresponding interval. At t = 0, v = 0, hence final velocity at t = 3s will increase to 6m/s. In next 3s, the velocity will decrease to zero. Thus, the velocity-time graph is
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A ball is thrown vertically up with a velocity of 20m/s. Construct acceleration time and displacement time graph.
Answer$\mathrm{u}=20 \mathrm{~ms}^{-1}, \mathrm{a}=\mathrm{gms}^{-2}$ Time to reach the highest point, $\text{t}=\frac{\text{u}}{\text{g}}=2\text{ seconds}.$ Max. height $=\frac{1}{2}\text{g}\times4=20\text{m}$
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In which of the following examples of motion, can the body be considered approximately a point object: A railway carriage moving without jerks between two stations.
AnswerThe size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
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The position$-$time$-$graph in figure depicts the journey of three bodies $A, B$ and $C.$
- At $1s$, which has the greatest velocity?
- At $2s$, which has travelled the farthest?
- When $A$ meets $C,$ is $B$ moving faster or slower than $A?$
- Is there any time at which the velocity of $A$ is equal to that of $B?$
Answer
- $B$
- $C$
- Slower
- Yes, in the interval $2$ to $3s$
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The velocity of a particle is given by equation $v=4+2\left(c_1+c_2 t\right)$, where $c_1$ and $c_2$ care constant. Find the initial velocity and acceleration of the particle.
AnswerGiven equation of velocity, $v=4+2\left(c_1+c_2 t\right) \Rightarrow v=\left(4+2 c_1\right)+2 c_2 t$ Compare the above equation with equation of motion $\mathrm{v}=\mathrm{u}+$ at Initial velocity, $\mathrm{u}=4+2 \mathrm{c}_1$ Acceleration of the Particle $=2 \mathrm{c}_2$.
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A body moving with a uniform acceleration describes $12m$ in $3^{rd}$ second of its motion and $20m$ in the $5^{th}$ second. Find the velocity after $10$ seconds.
AnswerLet the initial velocity of the body 'u' and acceleration 'a'. Using equation for $n^{th}$ second motion $\text{S}_{\text{n}^{\text{th}}}=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$
$\text{S}_{3^\text{rd}}=\text{u}+\frac{1}{2}\text{a}(2\times3-1)$
$12=\text{u}+\frac{5}{2}\text{a}\ \dots(\text{i})$
$\text{S}_{5^\text{th}}=\text{u}+\frac{1}{2}\text{a}(2\times5-1)$
$20=\text{u}+\frac{9}{2}\text{a}\ \dots(\text{ii})$ By solving equations (i) and (ii), we get $a = 4ms^{-2}$ and $u = 2ms^{-1}$ Now, Using equation $v = u +$ at v after 10 second = $2 + 4 \times 10ms v = 42ms^{-1}$.
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Two particles A and B are moving along the same straight line. B is ahead of A. Velocities remaining unchanged, what would be the effect on the magnitude of relative velocity if A is ahead of B?
AnswerThere will be no effect on the magnitude of relative velocity because relative velocity is $(\vec{\text{v}}_{\text{A}}-\vec{\text{v}}_{\text{B}})$ which always remains constant.
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Two bodies of different masses m, and m, are dropped from two different heights 'a' and 'b'. What is the ratio of time taken by the two to drop through these distances?
Answer$\text{a}=\frac{1}{2}\text{gt}^2_1$ $\text{t}_1=\sqrt{\frac{2\text{a}}{\text{g}}}$ $\text{b}=\frac{1}{2}\text{gt}^2_2$ $\text{t}_2=\sqrt{\frac{2\text{b}}{\text{g}}}$ $\therefore \frac{\text{t}_1}{\text{t}_2}=\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}}$
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The driver of a truck travelling with a velocity v suddenly notices a brick wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into the wall? Why?
Answer$\text{F}_\text{B}\times\text{d}=\frac{1}{2}\text{mv}^2$ $\text{F}_\text{B}=\frac{\text{mv}^2}{2\text{d}}$ $\text{F}_\text{T}=\frac{\text{mv}^2}{\text{d}}$ $\therefore \text{F}_\text{T}=2\text{F}_\text{B}$
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It is a common observation that rain clouds can be at about a kilometre altitude above the ground.A typical rain drop is about 4mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
AnswerKey concept: This problem can be solved by kinematic equations of motion and Newton's second law that $\mathrm{F}_{\mathrm{ext}}=\frac{\mathrm{dp}}{\mathrm{dt}}$ will be used, where dp is change in momentum over time dt . Diameter of the $\mathrm{drop}(\mathrm{d})=2 \mathrm{r}=4 \mathrm{~mm}$ Radius of the drop $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$ Mass of a rain $\operatorname{drop}(\mathrm{m})=\mathrm{v} \times \mathrm{p}$ $=\frac{4}{3} \pi \mathrm{r}^3 \mathrm{p}=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3\left(\because\right.$ Density of water $\left.=10^3 \mathrm{~kg} / \mathrm{m}^2\right) \Rightarrow \mathrm{m}=3.4 \times 10^{-5} \mathrm{~kg}$ Momentum of the rain $\operatorname{drop}(\mathrm{p})=\mathrm{mv}=3.4 \times 10^{-5} \times 100 \sqrt{2} \Rightarrow \mathrm{p}=4.7 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=5 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s}$
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If position of a particle at instant t is given by $\text{x}=\text{t}^3$, find acceleration of the particle.
AnswerGiven, $\text{x}=\text{t}^3$ $\therefore \nu=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{t}^3)=3\text{t}^2$ Now, acceleration (a) $=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3\text{t}^2)=6\text{t}.$
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Write two important points to distinguish displacement from distance.
AnswerLength of actual path covered between the initial and final points is distance while the length of the shortest path between initial and final points is displacement. The magnitude of displacement can be both positive and negative while distance is always positive.
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The velocity of a particle is given by equation $v=4+2\left(C_1+C_2 t\right)$ where, $C$, and $C$, are constant. Find the initial velocity and acceleration of the particle.
AnswerThe given equation is $v=4+2\left(C_1+C_2 t\right) \Rightarrow v=\left(4+2 C_1\right)+2 C_2 t$ Comparing the above equations with equation of motion $\mathrm{v}=\mathrm{u}+$ at Initial velocity, $\mathrm{u}=4+2 \mathrm{C}_1$ Acceleration of the particle $=2 \mathrm{C}_2$
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A body starts from rest and moves along a straight line. It has uniformly accelerated motion upto time $t_1$. During the interval $t_2-t_1$ it moves with uniform velocity. After time $t_2$ its motion is retarded, and it comes to rest at time $t_3$. Draw the velocity-time graph.
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The odometer of Raja's car reads 1700km at the start of a trip and 2500km at the end of the trip. The trip took 16h. What is the average speed of Raja's car in $ms^{-1}$?
AnswerHere, distance covered by the car = (2500 -1700)km = 800km Time elapsed = 16h Average speed $=\frac{\text{Total distance}}{\text{Total time}}$ $\text{u}=\frac{800}{16}=50\text{km/h}$ As 1km = 1000m and 1h = 3600s $\text{u}=\frac{50\times1000}{1\times3600}=13.9\text{ms}^{-1}$
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The velocity of a particle is $v=5+2\left(a_1+a_2 t\right)$ where $a_1$ and $a_2$ are constants and $t$ is the time. What is the acceleration of the particle?
Answer$\text{v}=5+2(\text{a}_1+\text{a}_2\text{t})$ $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[5+2(\text{a}_1+\text{a}_2\text{t})]$ $\text{a}=2\text{a}_2$
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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 velocity maximum?
AnswerGiven velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ For maximum velocity $\frac{\text{dv(t)}}{\text{dt}}=0$ $\Rightarrow\frac{\text{d}}{\text{dt}}(6\text{t}-2\text{t}^2)=0$ $\Rightarrow6-4\text{t}=0$ $\Rightarrow\text{t}=\frac{6}{4}=\frac{3}{2}\text{s}=1.5\text{s}$
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Points P, Q and R are in a vertical line such that PQ = QR. A ball at P is allowed to fall freely. What is the ratio of the times of descent through PQ and QR?
AnswerLet t, and tą be the times of descent through PQ and QR, respectively. Let PQ = QR = h Then, $\text{h}=\frac{1}{2}\text{gt}^2_1$ and $2\text{h}=\frac{1}{2}\text{g}(\text{t}_1+\text{t}_2)^2$ By dividing, we get $\frac{1}{2}=\frac{\text{t}^2_1}{(\text{t}_1+\text{t}_2)^2}$ $\frac{1}{\sqrt{2}}=\frac{\text{t}_1}{\text{t}_1+\text{t}_2}$ Hence, $\text{t}_1:\text{t}_2=2:(\sqrt{2}-1)$
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Find the area bounded by the curve $y = e^{-x}$, the X-axis and the Y-axis.
AnswerThe given function is $y = e^{-x}$ When $x = 0, y = e^{-0} = 1$ x increases, y value deceases and only at $\text{x}=\infty,\text{y} = 0.$ So, the required area can be found out by integrating the function from 0 to $\infty.$ So, $\text{Area}=\int\limits^{\infty}_0\text{e}^{\text{-x}}\text{dx}=-[\text{e}^{-\text{x}}]^{\infty}_0=1.$
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Two trains of lengths 109 m and 91 m are moving in opposite directions with velocities $34 \mathrm{~km} \mathrm{~h}^{-1}$ and $38 \mathrm{~km} / \mathrm{h}^{-1}$ respectively. In what time the two trains will completely cross each other? Choose the most logical reference point for time measurement.
AnswerRelative speed $=(34+38) \mathrm{km} / \mathrm{h}^{-1}=72 \mathrm{~km} / \mathrm{h}^{-1}=72 \times \frac{5}{18} \mathrm{~ms}^{-1}=20 \mathrm{~ms}^{-1}$ Total distance $=(109+91) \mathrm{m}=200 \mathrm{~m}$ Time $=\frac{200 \mathrm{~m}}{20 \mathrm{~ms}^{-1}}=10 \mathrm{~s}$
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Explain the difference between uniform velocity and variable velocity.
AnswerIf a body travels equal displacements in equal intervals of time, then the velocity of body is uniform velocity. On the other hand, if the body covers unequal displacements in equal intervals of time, then its velocity is variable velocity.
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Is earth inertial or non-inertial frame of reference?
AnswerSince, earth revolves around the sun and also spins about its own axis, so it is an accelerated frame of reference. Hence, earth is a non-inertial frame of reference. However, if we do not take large scale motion such as wind and ocean currents into consideration, we can say that approximation the earth is an inertial frame.
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Is the time variation of position shown in figure observed in nature possible?
AnswerNo, when x increases, the time first increases and then decreases. It is not possible. It implies that at a given time, the body in motion is simultaneously at two different positions which is not possible.
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A particle starts from rest, and its acceleration (a) plotted against time (t) is shown here. Plot the corresponding velocity (v) against time (t). Also plot the corresponding displacement (s) against time (t).
Answera = constant $\text{v}\propto \text{t}$ $\text{s}\propto \text{t}^2$ The corresponding graphs are: 
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The data regarding the motion of two different objects Pand Q are given in the following table. Examine them carefully and state whether the motion of the objects is uniform or non-uniform.
|
Time
|
Distance travelled by object P (in m)
|
Distance travelled by object Q (in m)
|
|
9:30 am
|
10
|
12
|
|
9:45 am
|
20
|
19
|
|
10:00 am
|
30
|
23
|
|
10:15 am
|
40
|
35
|
|
10:30 am
|
50
|
37
|
|
10.45 am
|
60
|
41
|
|
11:00 am
|
70
|
44
|
AnswerWe can see that the object P covers a distance of 10m in every 15 min. In other words, it covers equal distance in equal intervals of time. So, the motion of object Pis uniform. On the other hand, the object Q covers 7 m from 9:30 am to 9:45 am, 4 m from 9:45 am to 10:00 am and so on. In other words, it covers unequal distances in equal intervals of time. So, the motion of object Q is non-uniform.
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Explain how an object could have zero average velocity but non-zero average speed?
AnswerAverage velocity, $\text{v}_{\text{av}}=\frac{\text{Net displacement}}{\text{Total time taken}}$ and average speed, $\text{s}_{\text{av}}=\frac{\text{Total distanve travelled}}{\text{Total time taken}}$ If an object moves along a straight line starting from origin and then returns back to origin. Average velocity = 0 And Average Speed $=\frac{2\text{s}}{\text{t}}$
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The distance covered by a body is found to be directly proportional to the square of time. Is the body moving with uniform velocity or uniform acceleration? If the distance travelled be directly proportional to time.
AnswerLet $\text{s}\propto \text{t}^2$ $\text{s}=\text{kt}^2$ $\text{v}=\frac{\text{ds}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{kt})^2=2\text{kt},$ where k is costant and $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{v})$ $=\frac{\text{d}}{\text{dt}}(2\text{kt})=2\text{k}$ $\therefore$ The body is moving with uniform acceleration. In case $\text{s}\propto \text{t, s}= \text{kt}$ or $\text{v}=\frac{\text{ds}}{\text{dt}}=\text{k}$ (constant) and the body is moving with uniform velocity.
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What are uses of a velocity$-$time graph?
AnswerFrom a velocity$-$time graph, we can find out:
- The velocity of a body at any instant.
- The acceleration of the body and
- The net displacement of the body in a given time-interval.
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The position of an object moving along $x$-axis is given by $x=a+b t^2$ where $a=8.5 m , b=2.5 m s ^{-2}$ and $t$ is measured in seconds. What is its velocity at $t=0 s$ and $t=2.0 s$. What is the average velocity between $t=2.0 s$ and $t=4.0 s$ ?
AnswerIn notation of differential calculus, the velocity is
$
v=\frac{d x}{d t}=\frac{d}{d t}\left(a+b t^2\right)=2 b t=5.0 t m s ^{-1}
$
At $t=0 s , \quad V=0 m s ^{-1}$ and at $t=2.0 s$, $V=10 m s ^{-1}$.
Average velocity $=\frac{x(4.0)-x(2.0)}{4.0-2.0}$
$\begin{aligned}=\frac{a+16 b-a-4 b}{2.0} & =6.0 \times b \\ =6.0 \times 2.5 & =15 m s ^{-1}\end{aligned}$
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