Chapter 3 Motion in a Straight Line Questions and Answers: NCERT Solutions for Class 11th Physics (Partnership Physics )

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Class 11 Physics (Partnership Physics ) Chapter 3: Motion in a Straight Line - Questions and Answers of NCERT Book Solutions.

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1. In which of the following examples of motion, can the body be considered approximately a point object:

(a) a railway carriage moving without jerks between two stations.
Ans: As the size of a carriage is very small as compared to the distance between two stations, the carriage can be treated as a point sized object.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
Ans: As the size of a monkey is very small as compared to the size of a circular track, the monkey can be considered as a point sized object on the track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
Ans: As the size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground, the cricket ball cannot be considered as a point object.
(d) a tumbling beaker that has slipped off the edge of a table.
Ans: As the size of a beaker is comparable to the height of the table from which it slipped, the beaker cannot be considered as a point object

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2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below:

a. (A/B) lives closer to the school than (B/A)
Ans: From the graph, it is clear that OP OQOPOQ. Therefore, A lives closer to the school than B.
b. (A/B) starts from the school earlier than (B/A)
Ans: From the graph it is clear that for x = 0, t = 0
x = 0, t = 0
for A and t has some finite value for B. Hence, A starts from the school earlier than B.
c. (A/B) walks faster than (B/A)
Ans: As the velocity is equal to slope of x-t graph, in case of uniform motion and slope of x-t graph for B is greater than that for A. Thus B walks faster than A.
d. A and B reach home at the (same/different) time
Ans: From the graph it is clear that both A and B reach their respective homes at the same time.
e. (A/B) overtakes (B/A) on the road (once/twice).
Ans: As B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

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3. A woman starts from her home at 9.00 am, walks with a speed of 5 km/hr
5 km/hr
on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km/hr
25 km/hr
. Choose suitable scales and plot the x-t graph of her motion

Ans: In the above question it is given that:
Speed of the woman = 5 km/h
= 5 km/h
. Distance between her office and home = 2.5 km
= 2.5 km
. Time taken = Distance / Speed
Time taken = Distance / Speed
= 2.5 / 5 = 0.5 h = 30 min
= 2.5 / 5 = 0.5 h = 30 min
It is given that she covers the same distance in the evening by an auto.
Now, speed of the auto = 25 km/h
speed of the auto = 25 km/h
. Time taken = Distance / Speed
Time taken = Distance / Speed
= 2.5 / 25 = 1 / 10 = 0.1 h = 6 min
= 2.5 / 25 = 1 / 10 = 0.1 h = 6 min
The suitable x-t graph of the motion of the woman is shown in the given figure.

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4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Ans: In the above question it is given that:
Distance covered with 1 step = 1 m
= 1 m
Time taken = 1 s
= 1 s
Time taken to move first 5 m
5 m
forward = 5 s
= 5 s
Time taken to move 3 m
3 m
backward = 3 s
= 3 s
Net distance covered = 5 3 = 2 m
Net distance covered = 5 3 = 2 m
Net time taken to cover 2 m = 8 s
2 m = 8 s
. Hence,
Drunkard covers 2 m
2 m
in 8 s
8 s
. Drunkard covered 4 m
4 m
in 16 s
16 s
. Drunkard covered 6 m
6 m
in 24 s24 s.
Drunkard covered 8 m
8 m
in 32 s
32 s
. In the next 5 s
5 s
, the drunkard will cover a distance of 5 m 5 m
and a total distance of 13 m 13 m
and then fall into the pit.
Net time taken by the drunkard to cover kmkm13 m = 32 + 5 = 37 s
13 m = 32 + 5 = 37 s
The x-t graph of the drunkard’s motion can be shown below:

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5. A jet airplane travelling at the speed of 500 km/hr
500 km/hr
ejects its products of combustion at the speed of 1500 km/hr
1500 km/hr
relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans: In the above question it is given that:
Speed of the jet airplane, = 500 km/hr
= 500 km/hr
Relative speed of its products of combustion with respect to the plane,=1500 km/hr =1500 km/hr
. Speed of its products of combustion with respect to the ground =1500−500=1000km/hr=1500−500=1000km/hr
Hence, the speed of the latter with respect to an observer on the ground is 1000km/hr1000km/hr.

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6. A car moving along a straight highway with speed of 126 km/hr
126 km/hr
is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Ans: In the above question it is given that:
Initial velocity of the car is u=126 km/hr=35m/s
u=126 km/hr=35m/s
. Final velocity of the car is v=0km/hrv=0km/hr.
Distance covered by the car before coming to rest is 200 m
200 m
. Consider retardation produced in the car =a=a
From third equation of motion, v2−u2=2asv2−u2=2as
Therefore, −352=2a(200)−352=2a(200)
a=−3.0625m/s2a=−3.0625m/s2
From first equation of motion, time (t) taken by the car to stop can be obtained as:
v=u+atv=u+at
0=35−(3.065)t0=35−(3.065)t
t=11.44st=11.44s
Hence, the retardation of the car (assumed uniform) is 3.0625m/s23.0625m/s2, and it takes 11.44s11.44s for the car to stop.

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7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/hr
72 km/hr
in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s−2 1 m s−2
. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans: In the above question it is given that:
For train A: Initial velocity is u = 72 km/h = 20 m/s
u = 72 km/h = 20 m/s
Time is t=50 s
t=50 s
. Acceleration, a1=0a1=0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (s1)(s1) covered by train A can be obtained as:
Ans: In the above question it is given that:
For train A: Initial velocity is u = 72 km/h = 20 m/s
u = 72 km/h = 20 m/s
Time is t=50 s
t=50 s
. Acceleration, a1=0a1=0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (s1)(s1) covered by train A can be obtained as:
s=ut+(1/2)at2s=ut+(1/2)at2
=20×50+0=1000m=20×50+0=1000m
For train B:
Initial velocity is u = 72 km/h = 20 m/s
u = 72 km/h = 20 m/s
Acceleration, a=1m/s2a=1m/s2
Time is t=50 s
t=50 s
. From second equation of motion, distance (sn)(sn) covered by train A can be obtained as:
sn=ut+(1/2)at2sn=ut+(1/2)at2
=20×50+(1/2)(1)(50)2=2250m=20×50+(1/2)(1)(50)2=2250m
Hence,
Length of both trains = 2 × 400 m = 800 m
Length of both trains = 2 × 400 m = 800 m
Therefore, the original distance between the driver of train A and the guard of train B is 2250−1000 − 800 = 450m.

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8. On a two-lane road, car A is travelling with a speed of 36 km/hr
36 km/hr
. Two cars B and C approach car A in opposite directions with a speed of 54 km/hr
54 km/hr
each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Answer: (a) Wrap the thread a number of times on a round pencil so as to form a coil having its turns touching each other closely. Measure the length of this coil, mode by the thread, with a metre scale. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i.e., the thickness of the thread = 1/n .
This is equal to the diameter of the thread.
(b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, least count is decreased. Hence the accuracy is increased. However, this is only a theoretical
idea.Practically speaking, increasing the number of ‘turns would create many difficulties.
As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length.
(c) Due to random errors, a large number of observation will give a more reliable result than smaller number of
observations. This is due to the fact that the probability (chance) of making a positive random error of a given
magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of
observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtaine

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9. Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/hr
20 km/hr
in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

Ans: In the above question it is given that:
Consider V to be the speed of the bus running between towns A and B.
Speed of the cyclist is v=20 km/hr
v=20 km/hr
. Relative speed of the bus moving in the direction of the cyclist will be V−v=(V−20)m/sV−v=(V−20)m/s.
The bus went past the cyclist every 18
18
min i.e., 18 / 60 h
18 / 60 h
(when he moves in the direction of the bus).
Hence, distance covered by the bus = (V − 20) × 18 / 60 km
= (V − 20) × 18 / 60 km
...... (i) As one bus leaves after every T minutes, the distance travelled by the bus will be =V × T / 60
=V × T / 60
...... (ii)
As equations (i) and (ii) are equal.
(V − 20) × 18 / 60 = VT / 60
(V − 20) × 18 / 60 = VT / 60
...... (iii)
Relative speed of the bus moving in the opposite direction of the cyclist will be (V + 20) km/h (V + 20) km/h
. Thus, time taken by the bus to go past the cyclist = 6 min = 6 / 60 hr = 6 min = 6 / 60 hr
⇒(V + 20)×6 / 60 = VT / 60
⇒(V + 20)×6 / 60 = VT / 60
...... (iv)
From (iii) and (iv), we get
(V + 20) × 6 / 60 = (V − 20) × 18 / 60
(V + 20) × 6 / 60 = (V − 20) × 18 / 60
V + 20 = 3V 60
V + 20 = 3V 60
2V = 80
2V = 80
V = 40 km/h
V = 40 km/h
, which is the required speed.
Substituting the value of V in equation (iv),
(40 + 20) × 6 / 60 = 40T / 60
(40 + 20) × 6 / 60 = 40T / 60
T = 360 / 40 = 9 min
T = 360 / 40 = 9 min
, which is the required time period.

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10. A player throws a ball upwards with an initial speed of 29.4 m/s 29.4 m/s
a. What is the direction of acceleration during the upward motion of the ball?

Ans: Acceleration of the ball (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth, irrespective of the direction of the motion of the ball.
b. What are the velocity and acceleration of the ball at the highest point of its motion?
Ans: 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.,ng = 9.8 m/s2
g = 9.8 m/s2 . At maximum height, velocity of the ball becomes zero.
c. Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
Ans: The sign of position is positive, sign of velocity is negative, and sign of acceleration is positive during upward motion. During downward motion, the signs of position, velocity, and acceleration are all positive.
d. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m/s2 g = 9.8 m/s2
and neglect air resistance).
Ans: Initial velocity of the ball, u = 29.4 m/s u = 29.4 m/s
. Final velocity of the ball, v = 0
v = 0
(At maximum height, the velocity of the ball becomes zero)
Acceleration, a=g =9.8 m/s2
a=g =9.8 m/s2
From first equation of motion,
v=u+atv=u+at
0=−29.4+(9.8)t0=−29.4+(9.8)t
t=29.49.8=3st=29.49.8=3s
Time of ascent = Time of descent
Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands= 3 + 3 = 6 s = 3 + 3 = 6 s .

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11. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion
a. with zero speed at an instant may have non-zero acceleration at that instant

Ans: The above statement is true. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. It has acceleration equal to the acceleration due to gravity (g) Which acts in the downward direction at that point.
b. with zero speed may have non-zero velocity,
Ans: The above statement is false as speed is the magnitude of velocity. If speed is zero, the magnitude of velocity along with the velocity is zero.
c. with constant speed must have zero acceleration,
Ans: The above statement is true. If a car is moving on a straight highway with constant speed, it will have constant velocity. Acceleration is defined as the rate of change of velocity. Hence, the acceleration of the car is also zero.
d. with a positive value of acceleration must be speeding up.
Ans: The above statement is false. If acceleration is positive and velocity is negative at the instant time is taken as origin. Thus, for all the time before velocity becomes zero, there is slowing down of the particle. This case occurs when a particle is projected upwards. This statement will be true when both velocity and acceleration are positive, at that instant time taken as origin. This case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

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12. A ball is dropped from a height of 90 m 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 12 s t = 0 to 12 s .

Ans: In the above question it is given that:
Ball is dropped from a height is s = 90 m
s = 90 m
. Initial velocity of the ball is u = 0
u = 0
. Acceleration is a=g=9.8m/s2a=g=9.8m/s2.
Consider,
Final velocity of the ball to be v.
Using second equation of motion, time (t) taken by the ball to hit the ground can be obtained
as:
s=ut+(1/2)at2s=ut+(1/2)at2
90=0+(1/2)9.8t290=0+(1/2)9.8t2
t=18.38−−−−√=4.29st=18.38=4.29s
Using, first equation of motion, final velocity is given as:
v=u+atv=u+at
v=0+9.8(4.29)=42.04m/sv=0+9.8(4.29)=42.04m/s
Rebound velocity of the ball is calculated as:
ur=9v/10=9(42.0410)=37.84m/sur=9v/10=9(42.0410)=37.84m/s
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v=ur+at′v=ur+at′
0=37.84+(−9.8)t′0=37.84+(−9.8)t′
t′=37.84/9.8=3.86st′=37.84/9.8=3.86s
Total time taken by the ball will be t+t′=4.29+3.86=8.15st+t′=4.29+3.86=8.15s.
As the time of ascent is equal to the time of descent, the ball takes 3.86 s
3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor will be ur=9v/10=9(37.8410)=34.05m/sur=9v/10=9(37.8410)=34.05m/s
Total time taken by the ball for second rebound will be t+t′=8.15+3.86=12.01st+t′=8.15+3.86=12.01s
The speed-time graph of the ball is represented in the given figure as:

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13. Explain clearly, with examples, the distinction between:
a. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

Ans: The shortest distance (which is a straight line) between the initial and final positions of the particle gives the magnitude of displacement over an interval of time. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle is AC

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14. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/hr
5 km/hr
. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/hr
7.5 km/hr
.What is the (a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? (Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero)

Ans: In the above question it is given that:
Time taken by the man to reach the market from home is t1=2.5/5=1/2hr=30mint1=2.5/5=1/2hr=30min.
Time taken by the man to reach home from the market is t2=2.5/7.5=1/3hr=20mint2=2.5/7.5=1/3hr=20min.
Total time taken in the whole journey = 30 + 20 = 50 min
= 30 + 20 = 50 min
. 0 to 30 min
Average velocity = Displacement/Time
Average velocity = Displacement/Time
Average speed = Distance/Time
Average speed = Distance/Time
0 to 50 min
Time = 50 min = 50/60 = 5/6 h
Time = 50 min = 50/60 = 5/6 h
Net displacement = 0
Net displacement = 0 Total distance = 2.5 + 2.5 = 5 km
Total distance = 2.5 + 2.5 = 5 km
Average velocity = Displacement / Time = 0
Average velocity = Displacement / Time = 0
Average speed = Distance / Time = 5/(5/6) = 6 km/h
Average speed = Distance / Time = 5/(5/6) = 6 km/h
0 to 40 min
Speed of the man = 7.5 km/h
Speed of the man = 7.5 km/h
Distance travelled in first 30 min = 2.5 km
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
= 7.5 × 10/60 = 1.25 km
= 7.5 × 10/60 = 1.25 km
Net displacement = 2.5 1.25 = 1.25 km/
Net displacement = 2.5 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h
Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h
Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h
Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

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15. In Exercises 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? Guess where to put the missing c.

Ans: We know that instantaneous velocity is the first derivative of distance with respect to time.
Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. Therefore, both the total path length and magnitude of displacement become equal in this interval of time. Thus, instantaneous speed is always equal to instantaneous velocity.

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16. Look at the graphs (a) to (d) (figure) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Ans:
a. Consider the x-t graph, given in fig (a). It does not represent the one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
b. Consider the x-t graph, given in fig (b). It does not represent the one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
c. Consider the x-t graph, given in fig (c). It does not represent the one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
d. Consider the x-t graph, given in fig (d). It does not represent the one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

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17. Figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t<0t<0 and on a parabolic path for t>0t>0? If not, suggest a suitable physical context for this graph.

Ans: No, this is because the x-t graph does not represent the trajectory of the path followed by a particle. Also from the graph, it is clear that at t=0, x=0 t=0, x=0

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18. A police van moving on a highway with a speed of 30 km/hr
30 km/hr
fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/hr 192 km/hr
. If the muzzle speed of the bullet is 150 m/s 150 m/s
, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Ans: In the above question it is given that:
Speed of the police van is vp=30 km/hr=8.33m/s
vp=30 km/hr=8.33m/s
. Muzzle speed of the bullet is vb=150 m/s vb=150 m/s
. Speed of the thief’s car is vt=192 km/hr=53.33m/s vt=192 km/hr=53.33m/s
. As the bullet is fired from a moving van, its resultant speed will be: 150+8.33=158.33m/s150+8.33=158.33m/s.
Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:
vbt=vb−vt=158.33−53.33=105m/s

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19. Suggest a suitable physical situation for each of the following graphs (figure):

Ans: Consider fig 3.22 given in the question:
a. From the x-t graph given it is clear that initially a body was at rest. Further, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Further, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately stops after sometime.
b. From the given v-t graph it is clear that the sign of velocity changes and its magnitude decreases with a passage of time. This type of 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. c. From the given a-t graph it is clear 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 shows that the body again starts moving with the same constant velocity. This type of physical situation arises when a hammer moving with a uniform velocity strikes a nail.

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20. Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion.
(You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t=0.3s,1.2s,−1.2st=0.3s,1.2s,−1.2s .

Ans: Negative, Negative, Positive
Positive, Positive, Negative
Negative, Positive, Positive
When a particle executes simple harmonic motion (SHM), acceleration (a) is given by the relation: a=−ω2xa=−ω2x
Where,
ωω is the angular frequency ...
i. t = 0.3 s
t = 0.3 s
For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
ii. t =1.2s
t =1.2s
For this time interval, x is positive. Hence, the slope of the x-t plot will be positive. Thus, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
iii. t =−1.2s
t =−1.2s For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both x and t are negative, the velocity comes to be positive. From equation (i), it can be interpreted that the acceleration of the particle will be positive.

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21. Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Ans: The average speed is greatest in interval 3 and least in interval 2. It is positive in intervals 1 & 2 and negative in interval 3.
The average speed of a particle shown in the x-t graph is given by the slope of the graph in a particular interval of time.
From the graph it is clear that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Thus, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

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22. Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Ans: From the graph given in the question,
Average acceleration is greatest in interval 2
2 . Average speed is greatest in intervals of 33.
v is positive for intervals 1, 2
1, 2
, and 33.
a is positive for intervals 11 and 33 and negative in interval 22
a = 0
a = 0
at A, B, C, D
Acceleration is calculated as the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.
As the slope of the given speed-time graph is maximum in interval 2
2 , average acceleration will be the greatest in this interval.
From the time-axis, height of the curve gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Thus, average speed of the particle is the greatest in the interval 3.
For interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
As slope of the speed-time graph is negative, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
In interval 3:
As the slope of the speed-time graph is zero, acceleration is zero in this interval.
However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Thus, the slope is zero at these points.
Therefore, at points A, B, C, and D, acceleration of the particle is zero.

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23. A three-wheeler starts from rest, accelerates uniformly with 1 m/s2
1 m/s2
on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3....) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Ans: In the above question it is given that:
Initial velocity=u=0m/s
Initial velocity=u=0m/s
Acceleration=a=1m/s2
Acceleration=a=1m/s2
Distance covered by a body in nth second is given by the relation
sn=u+(2n−1)a/2sn=u+(2n−1)a/2
Hence,
sn=(2n−1)/2=n−1/2sn=(2n−1)/2=n−1/2 …… (1)
In this case, this relation shows that: snsn linearly varies with n.
Now, substituting different values of n in equation (1), we get the following table:

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24. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s
49 m/s
. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5m/s5m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Ans: In the above question it is given that:
Initial velocity of the ball is u = 49 m/s
u = 49 m/s . Acceleration is a=−g=−9.8m/s2a=−g=−9.8m/s2.
Consider,
case I:
When the lift was stationary, the boy throws the ball. Taking upward motion of the ball,
Final velocity, v of the ball becomes zero at the highest point.
From first equation of motion, time of ascent (t) is given as:
v=u+atv=u+at
t=v−ua=−49−9.8=5st=v−ua=−49−9.8=5s
However, the time of ascent is equal to the time of descent.
Thus, the total time taken by the ball to return to the boy’s hand is 5+5=10s5+5=10s.
Case II:
The lift was moving up with a uniform velocity of 5 m/s. Here, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s
49 m/s
. Therefore, in this case also, the ball will return back to the boy’s hand in 10 s
10 s .

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25. On a long horizontally moving belt (figure), a child runs to and fro with a speed 9 km/hr 9 km/hr
(with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h1
4 km h1

For an observer on a stationary platform outside, what is the
a. speed of the child running in the direction of motion of the belt?

Ans: As the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) will be:
vbB = vb + vB = 9 + 4 = 13 km/h
vbB = vb + vB = 9 + 4 = 13 km/h
. b. speed of the child running opposite to the direction of motion of the belt?
Ans: As the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + (−vB )= 9−4 = 5km/h
vbB = vb + (−vB )= 9−4 = 5km/h
c. time taken by the child in (a) and (b)? Which of the answers alter if motion is viewed by one of the parents?
Ans: We know that:
Distance between the childs parents =50m
Since, both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s
9 km/h = 2.5 m/s
. Thus, the time taken by the child to move towards one of his parents is 50/2.5 = 20s 50/2.5 = 20s
. If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h 9 km/h
. Therefore, it can be interpreted that the time taken by the child to reach any one of his parents remains unaltered.

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26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15m/s15m/s and 30m/s30m/s. Verify that the graph shown in figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10m/s2
g = 10m/s2
. Give the equations for the linear and curved parts of the plot.

Answer: Total time = 100 years = 100 x 365 x 24 x 60 x 60 s
Error in 1 second=0.02/100 x 365 x 24 x 60 x 60
=6.34 x 10-12 s
.•. Accuracy of 1 part in 1011 to 1012.

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27. The speed-time graph of a particle moving along a fixed direction is shown in figure. Obtain the distance traversed by the particle between (a)0s to 10s(a)0s to 10s, (b) 2s to 6s. What is the average speed of the particle over the intervals in (a) and (b)?

Ans: From the graph given in the question:
(a) Distance travelled by the particle = Area under the given graph
Distance travelled by the particle = Area under the given graph
=(1/2)×(10−0)×(12−0)=60m=(1/2)×(10−0)×(12−0)=60m
Average speed=Distance/Time=60/10=6m/s
Average speed=Distance/Time=60/10=6m/s
(b) Let s1M
s1
and s2
s2
be the distances covered by the particle between time t = 2 s
t = 2 s
to 5 s
5 s
and t = 5 s
t = 5 s
to 6 s
6 s
respectively.
Total distance (s) covered by the particle in time t = 2 s
t = 2 s
to 6 s
6 s
will be:
s=s1+s2s=s1+s2 …… (i)
For distance s1
s1
:
Let u′ be the velocity of the particle after 2 s
2 s
and a’ be the acceleration of the particle for t=0s to 5st=0s to 5s.
Since the particle undergoes uniform acceleration in the interval t=0s to 5st=0s to 5s, from first equation of motion, acceleration can be obtained as:
v=u+atv=u+at
Where,
v = Final velocity of the particle
v = Final velocity of the particle
a′=12/5=2.4m/s2a′=12/5=2.4m/s2
Again, from first equation of motion, we have
v=u+atv=u+at
=0+2.4×2=4.8m/s=0+2.4×2=4.8m/s
Distance travelled by the particle between time 2 s
2 s
and 5 s
5 s
i.e., in 3 s
3 s
will be:
s1=u′t+(1/2)a′t2s1=u′t+(1/2)a′t2
=4.8(3)+(1/2)×2.4×32=25.2m=4.8(3)+(1/2)×2.4×32=25.2m …… (ii)
For distance s2s2,
Let a’’ be the acceleration of the particle between time t=5s and t=10st=5s and t=10s.
From first equation of motion,
v=u+atv=u+at
0=12+a′′×50=12+a″×5
a′′=−2.4m/s2a″=−2.4m/s2
Distance travelled by the particle in 1s (i.e., t = 5s
t = 5s
to 6s
6s
)
s2=u′′t+(1/2)a′′t2s2=u″t+(1/2)a″t2
=12×1+(1/2)(−2.4)×(1)2=10.8m=12×1+(1/2)(−2.4)×(1)2=10.8m …… (iii)
From equations (i), (ii), and (iii), we get:
s=25.2+10.8=36m
s=25.2+10.8=36m
Hence, Average speed = 36 / 4 = 9 m/s
Average speed = 36 / 4 = 9 m/s

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28. The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29. Which of the following formulae are correct for describing the motion of the particle over the time-interval t1 to t2
t1 to t2a. x(t2)=x(t1)+v(t1)(t2−t1)+(1/2)a(t2−t1)2x(t2)=x(t1)+v(t1)(t2−t1)+(1/2)a(t2−t1)2
b. v(t2)=v(t1)+a(t2−t1)v(t2)=v(t1)+a(t2−t1)
c. vaverage=(x(t2)−x(t1))/(t2−t1)vaverage=(x(t2)−x(t1))/(t2−t1)
d. aaverage=(v(t2)−v(t1))/(t2−t1)aaverage=(v(t2)−v(t1))/(t2−t1)
e. x(t2)=x(t1)+vaverage(t2−t1)+(1/2)aaverage(t2−t1)2x(t2)=x(t1)+vaverage(t2−t1)+(1/2)aaverage(t2−t1)2
where
x(t2)−x(t1)=area under the v-t curve bounded by the t-axis and the dotted line as shown

Ans: The correct relations for the motion of the particle are (c), (d) and, (f). The given graph has a non-uniform slope. Thus, the relations given in (a), (b), and (e) cannot describe the motion of the particle. Only the relations given in (c), (d), and (f) are correct equations of motion.

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Last Updated on: Mar 20, 2024