WBCHSE Class 11 Physics One-Dimensional Motion Short Answer Questions

One-Dimensional Motion Short Answer Type Questions

Question 1. A particle starts from rest with constant acceleration. It travels a distance of x in the first 10 s and a distance of y in the next 20 seconds. The relation between x and y is

1. y = x
2. y = 2x
3. y = 8x
4. y = 4x

Answer:

We know, s = ut + \(\frac{1}{2}\)at²; u = 0

When t = 10 s, s = x; then x = \(\frac{1}{2}\) x a x 100 = 50a

When t = 10 + 20 = 30 s, s = x + y;

then x + y = \(\frac{1}{2}\) x a x 900 = 450a

∴ y = (x + y)-x = 450a-50a = 400a = 8x

The option 3 is correct.

Question 2. Show that the instantaneous speed of a particle is equal to the slope of the distance-time graph.
Answer:

The slope of the distance-time graph is \(\frac{ds}{dt}\).

Again, if Δt is the time in which the displacement of the particle is Δs, then speed, \(\frac{\Delta s}{\Delta t}\)

Now, if Δt → 0, instantaneous speed = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t}\)

Question 3. The position-time relation of a moving particle is x = 2t-3t².

1. What is the maximum +ve velocity of the particle?
2. When does the velocity of the particle become zero? (x is in metres and t is in seconds)

Answer:

x = 2t – 3t²

∴ v = \(\frac{dx}{dt}\) = 2 -6t

1. As t less than nor equal to 0, so, the maximum positive velocity of the particle is 2 m/s (at t = 0).
2. v = 0, when 2 – 6t = 0

∴ t = \(\frac{2}{6}\) = \(\frac{1}{3}\) s

Question 4. What information do we get from the slope of the velocity-time graph?
Answer:

We get the acceleration.

Question 5. A ball is thrown vertically upward. For the motion of the ball till it returns to the ground, draw the

1. Height vs time graph
2. velocity vs time graph
3. Acceleration vs time graph
4. Velocity vs height graph

Answer:

Question 6. A particle moves in the xy-plane with a constant acceleration of 4 m · s^-2 in the direction making an angle of 60° with the x-axis. Initially, the particle is at the origin and its velocity is 5 m · s^-1 along the x-axis. Find the velocity and the position of the particle at t = 5s.
Answer:

According to the question, if the initial velocity of the particle along OA is u, then (given, u_x = 5 m/s)

u = u_xcos60° = 5 x \(\frac{1}{2}\) = \(\frac{5}{2}\) m/s

The velocity of the particle at t = 5 s,

v = u+at = \(\frac{5}{2}\) + 4 x 5 = 22.5 m/s

At the time, the distance travelled by the particle along OA.

s = \(u t+\frac{1}{2} a t^2=\frac{5}{2} \times 5+\frac{1}{2} \times 4 \times 5^2\)

= \(\frac{25}{2}+50=\frac{125}{2} \mathrm{~m}\)

If the coordinate of the particle is (x, y) at that time, then

x = \(\frac{125}{2} \cos 60^{\circ}=\frac{125}{2} \times \frac{1}{2}=\frac{125}{4}\)

and y = \(\frac{125}{2} \sin 60^{\circ}=\frac{125}{2} \times \frac{\sqrt{3}}{2}=\frac{125 \sqrt{3}}{4}\)

∴ Position of the particle at t = \(5 \mathrm{~s} is \left(\frac{125}{4}, \frac{125 \sqrt{3}}{4}\right)\).

Question 7. A balloon is rising upwards from rest with acceleration \(\frac{g}{8}\). A stone is dropped from the balloon when it is at height H. Show that the time by which the stone will touch the ground is \(2 \sqrt{\frac{H}{g}}\).
Answer:

If the velocity of the balloon is v at a height H, \(v^2=2 \frac{g_8}{} H \quad \text { or, } \quad v=\frac{1}{2} \sqrt{g H}\)

The stone is dropped when the balloon is at height H.

If we consider the downward direction to be positive, the initial velocity of the stone,

u = \(-v=-\frac{1}{2} \sqrt{g H}\)

So, if the stone touches the ground by time t,

H = \(-\frac{1}{2} \sqrt{g H} \cdot t+\frac{1}{2} g t^2 \quad \text { or, } g t^2-\sqrt{g H} \cdot t-2 H=0\)

∴ t = \(\frac{\sqrt{g H} \pm \sqrt{g H+8 g H}}{2 g}=\frac{\sqrt{g H} \pm 3 \sqrt{g H}}{2 g}\)

Neglecting the negative value of t,

t = \(\frac{\sqrt{g H}+3 \sqrt{g H}}{2 g}=\frac{4 \sqrt{g H}}{2 g}=2 \sqrt{\frac{H}{g}}\)

Question 8. A car at rest accelerates at a constant rate for some time, after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t second, find the maximum velocity attained.
Answer:

If t₁ is the time taken to reach the maximum velocity v,

v = 0 + αt₁ or, \(t_1=\frac{\nu}{\alpha}\)

After decelerating, if the car takes time \(t_2\) to come to rest,

0 = \(\nu-\beta t_2 \quad \text { or, } t_2=\frac{\nu}{\beta}\)

So, \(t=t_1+t_2=\frac{\nu}{\alpha}+\frac{\nu}{\beta}=\nu \frac{\alpha+\beta}{\alpha \beta}\)

∴ \(v=\frac{\alpha \beta}{\alpha+\beta} t\)

Question 9. The initial velocity of a particle is u and acceleration (f) is uniform. The final velocity and distance covered in the interval t are v and s respectively. Show that the velocity of the particle at half-distance is more than the velocity for half-time.
Answer:

⇒ \(\nu=u+f t \text { or, } t=\frac{\nu-u}{f}\)

Again, \(v^2=u^2+2 f s\) or, \(s=\frac{v^2-u^2}{2 f}\)

Velocity at half time, \(v_1=u+f \frac{t}{2}=u+f \frac{\nu-u}{2 f}=u+\frac{\nu-u}{2}=\frac{u+v}{2}\)

If the velocity at half-distance is \(v_2\), \(\nu_2^2=u^2+2 f \frac{s}{2}=u^2+2 f \frac{v^2-u^2}{4 f}\)

= \(u^2+\frac{u^2+v^2}{2}=\frac{u^2+v^2}{2}\)

∴ \(v^2-v_1^2=\frac{u^2+v^2}{2}-\frac{(u+v)^2}{4}\)

= \(\frac{1}{4}\left\{2\left(u^2+v^2\right)-(u+v)^2\right\}\)

= \(\frac{1}{4}\left\{u^2+v^2-2 u v\right\}\)

= \(\left(\frac{v-u}{2}\right)^2\); since it is a whole square, it is a positive quantity.

So, \(\nu_2^2-\nu_1^2>0\)

or, \(v_2>v_1\)

Question 10. The equation of displacement of a particle along the Xaxis is x = 40 + 12 t- t³. How much distance does it travel before stopping?

1. 16 m
2. 40 m
3. 56 m
4. 36 m

Answer:

x = 40 + 12t – t³ or, \(\frac{dx}{dt}\) = 12 – 3t²

After time t s, if the particle comes of rest, then 0 = 12 -3t² or, t² = 4

∴ t = 2s

The distance travelled by the particle is 2 s, x|_t=2 = 2 = 40 + 12 x 2-8 = 56 unit

The option 3 is correct.

Question 11. In the s-t graph, a particle with uniform acceleration at time t makes an angle 45° with the time axis. After one second it makes an angle of 60°. What is the acceleration of the particle?
Answer:

Initial velocity of the particle, u = tan45° = 1 unit

After time 1 s, the final velocity of the particle, v = tan 60° = √3 unit

As the particle moves with uniform acceleration (a), so

a = \(\frac{\text { change in velocity }}{\text { time }}\) = (√3 – 1) unit

Question 12. Which of the following figures cannot be a speed-time graph?

Answer:

Time does not lag behind.

The option 4 is correct.

Question 13. A ball is dropped from the top of a building while another is thrown horizontally at the same instant. Which ball will strike the ground first?
Answer:

Both the balls touch the ground simultaneously.

Question 14. A body is moving from rest with an acceleration a m/s², which is related to time t s, by a = 3t + 4. What will be its velocity in time 2s?
Answer:

Acceleration of the body,

a = \(\frac{dv}{dt}\) = (3t + 4) m/s² at

At t = 2 s, the velocity of the body,

v = \(=\int_0^2(3 t+4) d t=\left[\frac{3 t^2}{2}+4 t\right]_0^2=14 \mathrm{~m} / \mathrm{s}\)

Question 15. A car travelling on a straight road moves with a uniform velocity v₁ for some time and with uniform velocity v₂ for the next equal time. The average velocity of the car is

1. \(\sqrt{v_1 v_2}\)
2. \(\frac{1}{v_1}+\frac{1}{v_2}\)
3. \(\frac{1}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right)\)
4. \(\frac{\left(v_1+v_2\right)}{2}\)

Answer:

If the car travels with velocity v₁ for time t and then with velocity v₂ for time t, then the displacement in time 2t = v₁t+ v₂t = (v₁ + v₂)t

∴ Average velocity = \(=\frac{\left(v_1+v_2\right) t}{2 t}=\frac{v_1+v_2}{2}\)

The option 4 is correct.

Question 16. The displacement of a particle is directly proportional to the third power of time. What will be the nature of the acceleration of the particle?
Answer:

s = kt³, where k = constant

∴ \(\frac{ds}{dt}\) = 3kt² and acceleration, a = \(\frac{d^2s}{dt^2}\) = 6kt

hence, a ∝ t

So, the acceleration of the particle is directly proportional to time t.

Question 17. A bullet enters a block of wood with a velocity u. Its velocity decreases to v after going through a distance x inside. After covering a further distance y inside, the bullet stops. Prove that \(\frac{u}{v}=\sqrt{\frac{y+x}{y}}\).
Answer:

Retardation of the bullet inside the block, a = constant

According to the equation, v² = u² – 2as, v² = u² – 2 ax

and, \(0=u^2-2 a(x+y) \text { or, } 2 a=\frac{u^2}{x+y}\)

∴ \(v^2=u^2-\frac{u^2 x}{x+y}=u^2\left(1-\frac{x}{x+y}\right)=\frac{u^2 y}{x+y}\)

or, \(\frac{u^2}{v^2}=\frac{x+y}{y} \text { or, } \frac{u}{v}=\sqrt{\frac{y+x}{y}}\)

Question 18. The velocity (m · s^-1)time(s) graph of a body is a straight line inclined at an angle of 45° with the time axis. The acceleration (in m · s^-2 unit) of the body is

1. 1
2. \(\frac{1}{\sqrt{2}}\)
3. \(\sqrt{2}\)
4. \(\frac{1}{\sqrt{3}}\)

Answer:

If angle of inclination is θ, then acceleration, a = tanθ

∴ a = tan 45° = 1

The option 1 is correct.

Question 19. At any instant the speeds of two identical cars with the same retardation are u and 4u; starting from that instant the respective distances the cars travel before stopping are in the ratio

1. 1:1
2. 1:4
3. 1:8
4. 1:16

Answer:

The final velocity of both the cars, v = 0

Let the first and the second cars travel distances d₁ and d₂ before coming to rest. If the retardation of both the cars is a, then in the case of the first car,

⇒ \(-u^2=-2 a d_1 \text { or, } d_1=\frac{u^2}{2 a}\)

in the case of the second car, \(-(4 u)^2=-2 a d_2 \text { or, } d_2=\frac{8 u^2}{a}\)

∴ \(\frac{d_1}{d_2}=\frac{u^2}{2 a} \cdot \frac{a}{8 u^2}=\frac{1}{16}\)

Hence, \(d_1: d_2=1: 16\)

The option (4) is correct.

Question 20. The distance-time graph of a moving particle is given by x = 4t – 6t²

1. What is the positive maximum speed?
2. At what time would the speed of the particle be zero? (x is in metres and t is in seconds)

Answer:

Velocity of the particle, v = \(\frac{dx}{dt}\) = 4 – 12t

Acceleration of the particle, a = \(\frac{d^2 x}{dt^2}\) = -12 m/s²

So, the particle is moving with retardation.

At t = 0, the velocity is maximum. In this case,

1. The maximum positive velocity of the particle = 4- 12 x 0 = 4m/s
2. When v = 0; 4 – 12t = 0 or, t = \(\frac{1}{3}\)s

∴ The velocity of the particle will be zero after time \(\frac{1}{3}\) s.

Question 21. A particle moves with constant acceleration along a straight line starting from rest The percentage increase in its displacement during the 4th second compared to that in the 3rd second is

1. 33%
2. 40%
3. 66%
4. 77%

Answer:

⇒ \(s_n=u+\frac{1}{2} a(2 n-1)\)

⇒ \(s_3=\frac{5}{2} a, s_4=\frac{7}{2} a\)

∴ \(\frac{s_4-s_3}{s_3} \times 100=\frac{a}{\left(\frac{5 a}{2}\right)} \times 100=40 \%\)

The option 2 is correct.

Question 22. Two particles A and B having different masses are projected from a tower with the same speed. A is projected vertically upward and B is vertically downward. On reaching the ground

1. The velocity of A is greater than that of B
2. The velocity of B is greater than that of A
3. Both A and B attain the same velocity
4. The particle with the larger mass attains a higher velocity

Answer:

Motion under gravity is independent of mass. The particle A will come back to the tower with the same velocity at which it was thrown vertically upwards. Hence, the downward velocities of both particles A and B at the top of the tower are equal. Therefore, both the particles will attain the same velocity on reaching the ground.

The option 3 is correct

Question 23. At a particular height, the velocity of an ascending body is \(\vec{u}\). The velocity at the same height while the body falls freely is

1. 2\(\vec{u}\)
2. –\(\vec{u}\)
3. \(\vec{u}\)
4. -2\(\vec{u}\)

Answer:

While moving up and moving down, at a particular height, the velocity of a body is the same and in opposite directions.

The option 2 is correct

Question 24. A train moves from rest with acceleration α and in time t₁ covers a distance x. It then decelerates to rest at constant retardation β for distance y in time t₂. Then

1. \(\frac{x}{y}=\frac{\beta}{\alpha}\)
2. \(\frac{\beta}{\alpha}=\frac{t_1}{t_2}\)
3. \(x=y\)
4. \(\frac{x}{y}=\frac{\beta t_1}{\alpha t_2}\)

Answer:

In the first case, \(\nu=u+a t \quad \text { or, } \nu=0+\alpha t_1=\alpha t_1 \quad \text { or, } x=\frac{1}{2} \alpha t_1^2\)

In the second case, \(v^2=u^2-2 a s \text { or, } 0=\alpha^2 t_1^2-2 \beta y\) (because \(u=\alpha t_1\))

or, \(y=\frac{\alpha^2 t_1^2}{2 \beta}\)

∴ \(\frac{x}{y}=\frac{\frac{1}{2} \alpha t_1^2}{\frac{\alpha^2 t_1^2}{2 \beta}}=\frac{\beta}{\alpha}\)

Also, \(\nu=u-a t\) or, \(0=\alpha t_1-\beta t_2\) or, \(\frac{t_1}{t_2}=\frac{\beta}{\alpha}\)

The options (1) and (2) are correct.

Question 25. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is

1. 2gH = n²u²
2. gH = (n-2)²u²
3. 2gH= nu²(n-2)
4. gH = (n-2)²u²

Answer:

Time taken to reach the highest point = \(\frac{u}{g}\)

Speed on reaching the, ground = \(\sqrt{u^2+2 g h}\)

Now, v = u+ at

or, \(\sqrt{u^2+2 g h}=-u+g t\)

or, t = \(\frac{u+\sqrt{u^2+2 g H}}{g}\)

According to the question, \(\frac{u+\sqrt{u^2+2 g H}}{g}=\frac{n u}{g}\)

or, 2gH= n(n-2)u²

The option 3 is correct.

Question 26. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speeds of 10 m/s and 40 m/s respectively. Which of the following graphs best represents the time variation of the relative position of the second stone with respect to the first?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s²) (The figures are schematic and not drawn to scale)

Answer:

We consider the upward direction to be positive,

∴ y = ut – \(\frac{1}{2}\)gt²

Till they reach the ground, for the first stone,

-240 = 10t – \(\frac{1}{2}\) x 10 t²

or, 5t² – 10t – 240 = 0 or, 5(t+4)(t-8) = 0

∴ t = 8

for the second stone,

-240 = 40t – \(\frac{1}{2}\) x 10 x t²

∴ t = 12 s

So, for the last (12 – 8=)4s, only the second stone will be in motion.

Hence, in the time span between 8 s and 12 s, y₂ – y₁ = (40t-5t²) – 0 = 40t – 5t² which is the equation of a parabola.

Now, in the time span between 0 s to 8 s, y₂ – y₁ = (40t-5t²)-(10t-5t²) = 30t

which is the equation of a straight line passing through the origin.

The option 3 is correct.

Question 27. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

Answer:

Slope of the v-t graph indicates acceleration. Here acceleration due to gravity is constant.

The option 3 is correct

Question 28. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Answer:

Each of the graphs 1, 2 and 3 represents the motion of a body thrown vertically upward. But option 2 does not represent any motion like that.

Option 4 is correct.

Question 29. A particle Of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx^-2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by

1. \(-2 n \beta^2 x^{-2 n-1}\)
2. \(-2 n \beta^2 x^{-4 n-1}\)
3. \(-2 \beta^2 x^{-2 n+1}\)
4. \(-2 n \beta^2 e^{-4 n+1}\)

Answer:

Acceleration, a = \(\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{d v}{d x} v=\nu \frac{d v}{d x}\)

= \(\beta x^{-2 n}\left\{(-2 n) \beta x^{-2 n-1}\right\}=-2 n \beta^2 x^{-4 n-1}\)

The option 2 is correct.

Question 30. If the velocity of a particle is v = At+ Bt², where A and B are constant, then the distance travelled by it between 1s and 2s is

1. 3A + 7B
2. \(\frac{3}{2}A+\frac{7}{3}\)B
3. \(\frac{A}{B}+{B}{3}\)
4. \(\frac{3}{2}\)A + 4B

Answer:

The velocity of the particle, v = At+ Bt²

or, \(\frac{d s}{d t}=A t+B t^2\)

or, \(\int_0^s d s=\int_1^2\left(A t+B t^2\right) d t\)

or, \(s=\left[\frac{A t^2}{2}+\frac{B t^3}{3}\right]_1^2=2 A+\frac{8 B}{3}-\frac{A}{2}-\frac{B}{3}=\frac{3 A}{2}+\frac{7 B}{3}\)

∴ The distance travelled by it between 1s and 2s

= \(\frac{3 A}{2}+\frac{7 B}{3}\)

The option 2 is correct

Question 31. A car moving with a speed of 50 km · h^-1 can be stopped by brakes after at least 6 m. What will be the minimum stopping distance, if the same car is moving at a speed of 100 km · h^-1?
Answer:

Final velocity, v = 0; if u = initial velocity, a = retardation and s = distance travelled after applying the brakes, then, v² = u²-2as or, 0 = u²-2as

or, s = \(\frac{u^2}{2a}\). In this example, a is the maximum retardation produced by the brakes.

So, s is the minimum stopping distance.

Then, \(\frac{s_1}{s_2}=\left(\frac{u_1}{u_2}\right)^2 \text { or, } s_2=s_1\left(\frac{u_2}{u_1}\right)^2=6 \times\left(\frac{100}{50}\right)^2=24 \mathrm{~m}\)

Question 32. The displacement-time graphs of two bodies P and Q are represented by OA and BC respectively. What is the ratio of the velocities of P and Q? ∠OBC = 60° and ∠AOC = 30°

Answer:

Velocity, v = \(\frac{ds}{dt}\) = tanθ = slope of the displacement-time graph, where θ = angle made with the time axis.

∴\(\frac{v_P}{v_Q}=\frac{\tan \theta_P}{\tan \theta_Q}=\frac{\tan 30^{\circ}}{\tan \left(-30^{\circ}\right)}=\frac{\tan 30^{\circ}}{-\tan 30^{\circ}}=\frac{\frac{1}{\sqrt{3}}}{-\frac{1}{\sqrt{3}}}=-\frac{1}{1}\)

So the ratio is 1: (-1).

Question 33. What does the slope of a velocity-time graph represent?
Answer:

Let time t be plotted along the horizontal axis and velocity v along the vertical axis. The slope of this velocity-time graph is \(\frac{dv}{dt}\). Again, by definition, acceleration a = \(\frac{dv}{dt}\), so the slope represents acceleration.

Question 34. Draw a velocity-time graph for an object starting from rest. Acceleration is constant and remains positive.
Answer:

For an object starting from rest, v = 0 at t = 0. This is represented by the origin O of the velocity-time graph.

The slope of the graph = acceleration. As the acceleration is constant and positive, the straight line OA, having a constant and positive slope represents the motion.

Question 35. An object moving on a straight line covers the first half of the distance at speed v and the second half at speed 2 v. Find

1. Average speed,
2. Mean speed.

Answer:

1. Let total distance = D.

So, time taken to cover the first half = \(\frac{D / 2}{v}=\frac{D}{2 v}\)

and time taken to cover the second half =\(\frac{D / 2}{2 v}=\frac{D}{4 v}\)

Hence, the average speed total distance \(=\frac{\text { total distance }}{\text { total time }}\)

= \(\frac{D}{\frac{D}{2 v}+\frac{D}{4 v}}=\frac{D}{D\left(\frac{2+1}{4 v}\right)}=\frac{4 v}{3}\)

2. Mean speed \(=\frac{v+2 v}{2}=\frac{3 v}{2}\)

Question 36. A ball is thrown vertically upwards. Draw its

1. Velocity time graph,
2. Acceleration-time graph

Answer:

1. For the upward motion, the velocity decreases uniformly with retardation g, where g is the acceleration due to gravity. The line AB represents this motion. Then, for the downward motion, the velocity increases uniformly for the same time interval with acceleration g. The line BC same represents this motion.
2. The entire motion is under acceleration due to gravity (g), which is a constant. So the acceleration-time graph is a horizontal straight line.

Question 37. A car is moving along a straight line in the given. It moves from O to P in 18 seconds and returns from P to Q in 6 seconds. What are the average velocity and average speed of the car in going

1. From O to P? and
2. From O to P and back to Q?

Answer:

For the motion from O to P, distance travelled = displacement = OP = (360-0) = 360 m

So, average speed = \(\frac{\text { distance travelled }}{\text { time }}\)

= \(\frac{\text { displacement }}{\text { time }}=\text { average velocity }\)

This average speed or velocity is \(\nu=\frac{360 \mathrm{~m}}{18 \mathrm{~s}}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Now, for this motion from O to P and back to Q, distance travelled = OP+ PQ

= 360 + (360 – 240) = 480 m

So, average speed = \(\frac{480 \mathrm{~m}}{(18+6) \mathrm{s}}=20 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

But displacement = \(\overrightarrow{O P}+\overrightarrow{P Q}=\overrightarrow{O Q}\), whose magnitude = 240 m and the time taken = 18 + 6 = 24 s.

Hence, average velocity = \(\frac{240 m}{24 s}\) = 10 m · s^-1

Question 38. Draw

1. Position-time,
2. Velocity-time and
3. Acceleration-time graph for the motion of an object under free fall.

Answer:

We assume the downward direction as the positive direction during this free fall.

1. As u = 0, the displacement at any time t is h = 1/2 gt² or t²= 2/g h.

So the position-time graph is a parabola passing through the origin [like x² = 4 ay on an x-y graph].

2. Initial velocity u = 0, so, at any time t the velocity is v = gt.

Hence, the velocity-time graph is a straight line of positive slope passing through the origin [like y = mx on a xy-graph]

3. Acceleration (a) = acceleration due to gravity (g) = constant and positive.

So, the acceleration-time graph is parallel with the time axis.

Question 39. The x-t graph of an object in straight-line motion is shown. Predict the type of motion it undergoes.

Answer: It is a motion with uniform velocity, as the graph is a straight line

Question 40.

1. Having seen a big stone falling from the top of a tower Rabi pulled his friend Kiran away. The stone hits Kiran slightly and he gets a little hurt. But he was saved from a major accident. What made Rabi act in such a way?
2. From the top of a tower 100 m in height, a ball is dropped and at the same time, another ball is projected vertically upwards from the ground with a velocity of 25 m · s^-1. Find when and where the two balls meet. Take g = 9.8 m · s^-2.

Answer:

1. Rabi’s first observation that the heavy object was falling freely from the top of the tower, and then registering in his brain of the danger of the object hitting his friend Kiran—these are the reasons that made Ravi act so quickly.
   • Fortunately, his reaction time was just low enough to save Kiran. In this case, the reaction time (τ) is equal to the time of flight (t) of the object in a free fall from the height (h) of the top of the tower. So, Ravi’s reaction time can be estimated from the knowledge of h.
2. AB = height of the tower = 100 m

Let the two balls meet after a time t at point C, such that AC = h₁, BC = h₂ and h₁ + h₂ = 100 m.

For the first ball, h₁ = \(\frac{1}{2}{g t^2}\)

For the second ball, h₂ = ut – \(\frac{1}{2}{g t^2}\)

By adding, we get h₁ + h₂ = ut

or, 100 = 25t or, t = 4 s

So, the two balls meet after 4 s.

The height of point C, where the two balls meet, from the ground is, h₂ = ut – \(\frac{1}{2}{g t^2}\)

= 25 x 4 – \(\frac{1}{2}\) x 9.8 x 4² =100-78.4

Question 41. The displacement of a particle along the x-axis is given by x = 3 + 8t-2t². What is its acceleration? At what time it will come to rest? All are in SI units.
Answer:

Displacement, x = 3 + 8t – 2t²;

velocity, v = \(\frac{dx}{dt}\) = 8 – 4t

acceleration, a = \(\frac{dv}{dt}\) = -4 m · s^-2.

This negative value means that the particle is moving with a uniform retardation.

When the particle comes to rest, v = 0; i.e., 8 – 4t = 0,

∴ t = 2 s

Question 42. The acceleration-time graph for a body is shown. Plot the corresponding velocity-time graph and draw the inference. The body starts with non-zero positive velocity.

Answer:

The given figure shows that the acceleration a is a positive constant—it does not change with time. So, this represents a motion under uniform acceleration.

Then, on the velocity-time graph, motion is given by a straight line of positive slope. The velocity increases uniformly from the initial value u to the final value v at time t.

Question 43. The position coordinate of a moving particle is given by x = 6 + 18t + 9t² (where x is in metres, t in seconds. What is its velocity and acceleration at t = 2s.
Answer:

Given position coordinate, x = 6 + 18t+ 9t² = 9t² + 18t+ 6

Differentiating the above equation with respect to t, \(\frac{dx}{dt}\) = 18t+ 18 or, v = 18t+ 18 dt ….(1)

Given, t = 2 s

Therefore the velocity at 2 s is v = 36 + 18 = 54 m/s

Again, differentiating equation (1) with respect to time t, \(\frac{dv}{dt}\)

Therefore the acceleration, a = 18 m/s²

Question 44. Is it possible to have a constant rate of change of velocity when velocity changes both in magnitude and direction?
Answer:

Yes, when a body moves upwards or downwards, the rate of change of velocity remains constant while the magnitude and direction of velocity change.

Question 45. The velocity of a moving particle is given by, v = 6 + 18t + 9t² (x in metre, t in second) what is its acceleration at t = 2 s?
Answer:

Given, u = 6 + 18t + t²

∴ Acceleration, a = \(\frac{dv}{dt}\) = 18 + 18t

Hence, acceleration at t = 2 s is \(\left.\frac{d v}{d t}\right|_{t=2}=18+18 \times 2=54 \mathrm{~m} / \mathrm{s}^2\)

Question 46. Plot the position-time graph for an object

1. Moving with positive velocity,
2. Moving with negative velocity and
3. At rest.

Answer:

Moving with positive velocity

Moving with negative velocity

At rest

Question 47. The position of an object moving along x -x-axis is given by x = a+bt² where a = 8.5 m, b = 2.5 m/s² and t is measured in seconds. What is its velocity at t = 0 and t = 2.0 s? What is the average velocity between t = 2.0 s and t = 4.0 s?
Answer:

Given, x = a+bt²

Now, instantaneous velocity, v = \(\frac{dx}{dt}\) = 2bt

So, at t=0, \(\left.\frac{d x}{d t}\right|_{t=0}=0\); at \(t=2 \mathrm{~s},\left.\frac{d x}{d t}\right|_{t=2}=4 b=10 \mathrm{~m} / \mathrm{s}\)

Now, at t=2s, \(x_1=a+4 b\) at t=4s, \(x_2=a+16 b\)

∴ Average velocity = \(\frac{x_2-x_1}{t_2-t_1}\) = \(\frac{(a+16 b)-(a+4 b)}{4-2}\)

= \(\frac{12 b}{2}=15 \mathrm{~m} / \mathrm{s}\) (because b=2.5)