NCERT solutions for Class 11 Physics Chapter 2: Motion in a Straight Line

(a) The railway carriage moving without jerks between two stations, so the distance between two stations is considered to be large as compared to the size of the train. Therefore the train is considered as a point object.

(b) The monkey may be considered as point object because value of distance covered on a circular track is much greater.

(c) As turning of the ball is not smooth, thus the distance covered by the ball is not large in the reasonable time. Therefore the ball cannot be considered as a point object.

(d) Again a tumbling beaker slipped off the edge of a table cannot be considered as a point object because distance covered is not much larger.

 (a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],
(b) A starts earlier for school than B, because t = 0 for A but for B, t has some finite time.

(c) As slope of B is greater than that of A, thus B walks faster than A.

(d) A and B reach home at the same time.

(e) At the point of intersection (i.e., X), B overtakes A on the roads once.

Distance covered while walking = 2.5 km.
Speed while walking = 5 km/h
Time taken to reach office while walking = (2.5/5 ) h=1/2 h

If O is regarded as the origin for both time and distance, then at t = 9.00 am, x = 0
and at t = 9.30 am, x = 2.5 km
OA is the x-t graph of the motion when the woman walks from her home to office. Her stay in the office from 9.30 am to 5.00 pm is represented, by the straight line AB in the graph.
Now, time taken to return home by an auto = 2.5/5 h =1/10 h =6 minute
So, at t = 5.06 pm, x = 0
This motion is represented by the straight line BC in the graph. While drawing the x-t graph, the scales chosen are as under:
Along time-axis, one division equals 1 hour.
Along positive-axis, one division equals 0.5 km.

Since the man steadily moves forward as the time progresses so the following graph will represent his motion till he covers 13 m. In 5 s he moves through a distance of 5 m and then in next 3 s comes back by 3 m.

Thus in 8 s he covers only 2 m, as shown in the graph he would fall in the pit in 37 s.

As pointed out earlier, the man covers 2 m in 8 s so, he will cover 8 m in 32 s. But at the end in 5 s he would cover another 5 m i.e., 32 s + 5 s = 37 s, he would cover 8 m + 5 m = 13 m. Thus, he would fall in the pit in 37th second.

Velocity of jet airplane w.r.t observer on ground = 500 km/h.
If Vj and v0 represent the velocities of jet and observer respectively, then vj – vo = 500 km h-1
Similarly, if vc represents the velocity of the combustion products w.r.t jet plane, then vc – vg = -1500 km/h

The negative sign indicates that the combustion products move in a direction opposite to that of jet.

Speed of combustion products w.r.t. observer

= vc – u0 = (vc – vj) + (vj – v0) = (-1500 + 500) km h-1 = -1000 km h-1.

 Here length of train A = length of train B = l = 400 m. As speed of both trains u = 72 km h-1 = 20 ms-1 in same direction, hence their relative velocity uBA = 0.
Let initial distance between the two trains be ‘S’ then train B covers the distance (S + 11) = (S + 800) m in time t = 50 s when accelerated with a uniform acceleration a = 1 m/s2.

and initial distance between guard of train B from driver of train A = 450 + 800 = 1250 m.

 Let vb be the speed of each bus. Let vc be the speed of cyclist.
Relative velocity of the buses plying in the direction of motion of cyclist is vb – vc .
The buses playing in the direction of motion of the cyclist go past him after every
18 minute i.e.18/20 h.

(a) The direction of acceleration during the upward motion of the ball is vertically downward.
(b) At the highest point, velocity of ball is zero but acceleration (g = 9.8 ms-2) in vertically downward direction.
(c) If we consider highest point of ball motion as x = 0, t = 0 and vertically downward direction to be +ve direction of x-axis, then
(i) during upward motion of ball before reaching the highest point position (as well as displacement) x = +ve, velocity v = -ve and acceleration a = g = +ve.
(ii) during the downward motion of ball after reaching the highest point, x, v and a = g all the three quantities are positive.
(d) During upward motion

 (a) True. Consider a ball thrown up. At the highest point, speed is zero but the acceleration is non-zero.
(b) False. If a particle has non-zero velocity, it must have speed.
(c) True. If the particle rebounds instantly with the same speed, it implies infinite acceleration which is physically impossible.
(d) False. True only when the chosen position direction is along the direction of motion.

 (a) Suppose a particle goes from point A to B along a straight path and returns to A along the same path. The magnitude of the displacement of the particle is zero, because the particle has returned to its initial position. The total length of path covered by the particle is AB + BA = AB + AB = 2 AB. Thus, the second quantity is greater than the first,
(b) Suppose, in the above example, the particle takes time t to cover the whole journey. Then, the magnitude of the average velocity of the particle over time-interval t is = Magnitude of displacement /Time-interval =0/t =0
While the average speed of the particle over the same time-interval is =Total path length /Time-interval= 2 AB /t
Again, the second quantity (average speed) is greater than the first (magnitude of average velocity).
Note: In both the above cases, the two quantities are equal if the particle moves from one point to another along a straight path in the same direction only.

Instantaneous velocity is the velocity of a particle at a particular instant of time. In this case of a small interval of time, the magnitude of the displacement is effectively equal to the distance travelled by the particle in the same interval of time. Therefore, there is no distinction between instantaneous velocity and speed.

None of the four graph represent a possible one-dimensional motion. In graphs (a) and (b) motions are definitely two dimensional. Graph (a) represents two positions at the same time which is not possible.
In graph (b) opposite motion is visible at the same time.
The graph (c) is not correct since it shows that the particle has negative speed at a certain instant. Speed is always positive.
In graph (d) path length is shown as increasing as well as decreasing. Path length never decreases.

 It is not correct to say that the particle moves in a straight line for t < 0 (i.e., -ve) and on a parabolic path for t > 0 (i.e., + ve)
because the x-t graph can not show the path of the particle.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t =0.

(a) A ball at rest on a smooth floor is kicked. It rebounds from a wall with reduced speed and moves to the opposite wall which stops it.
(b) The graph shows that velocity changes again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after thrown up), on striking the ground rebounds with reduced speed after each hit against the ground.
(c) A uniformly moving cricket ball turned back by hitting it with a bat for a very short time-interval.

 In x-t graph of Fig. showing simple harmonic motion of a particle, the signs of position, velocity and acceleration are as given below.

In S.H.M., acceleration, a – x or a = – kx.
(i) At t = 0.3 s, x < 0 i.e., x is in -ve direction. Moreover, as x is becoming more negative with time, it shows that v is also – ve (i.e., v < 0). However, a = -kx will be +ve (a > 0).
(ii) At t = 1.2 s, x > 0, v > 0 and a < 0.
(iii) At t = -1.2 s, x < 0, but here on increasing the time t, value of x becomes less negative.
It means that v is +ve (i.e., v > 0). Again a = – kx will be positive (i.e., a > 0).

Greater in 3, least in 2; v > 0 in 1 and 2, v < 0 in interval 3.

The acceleration is greatest in magnitude in interval 2 as the change in speed in the same time is maximum in this interval.
The average speed is greatest in interval 3 (peak D is at maximum on the speed axis).
The sign of v and a in the three intervals are:
v > 0 in 1, 2 and 3; a > 0 in 1
a < 0 in 2, a=0 in 3.
acceleration is zero at A,B,C and D.

When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinetnatic motion equations presuming a = ± g (as the case may be).In present case u = 49 ms-1 (upward) a = g = 9.8 ms-2 (downward)
If the ball returns to boy’s hands after a time t, then displacement of ball relative to boy
is zero i.e., s = 0. Hence, using equation s = ut + 1/2 at 2, we have

Speed of child with respect to belt = 9 km h-1 Speed of belt = 4 km h-1
(a) When the child runs in the direction of motion of the belt, then speed of child w.r.t. stationary observer = (9 + 4) km h-1 = 13 km h-1.
(b) When the child runs opposite to the direction of motion of the belt, then speed of child w.r.t. stationary observer = (9 – 4) km h-1 = 5 km h-1
(c) Speed of child w.r.t . either parent=9 km h-1
Distance to be covered = 50 m = 0.05 km 0.05 km
Time=0.05 km/9k h-1=0.0056 h =20 S
If the motion is viewed by one of the parents, then the answers to (a) and (b)are altered but answer to (c) remains unaltered.

For first stone,
x (0) = 200 m, v (0) = 15 ms-1, a = -10 ms-2
x1 (t) = x (0) + v (0) t + 1/2 a t2
x1 (t) = 200 + 15t – 5t2
When the first stone hits the ground, x1 (t) = 0
– 5t2 + 15t+ 200 = 0 On simplification, t = 8 s
For second stone, x (0) = 200 m, v (0) = 30 ms-1, a = -10 ms-2
x1 (t) = 200 + 30t – 5t2
When this stone hits the ground, x1(t) = 0 .-. -5t2 + 30t + 200 = 0
Relative position of second stone w.r.t. first is given by x2 (t) – x1 (t) = 15t
Since there is a linear relationship between x2(t) – x1 (t) and t, therefore the graph is a straight line.
For maximum separation, t = 8 s So maximum separation is 120 m
After 8 second, only the second stone would be in motion. So, the graph is in accordance with the quadratic equation.

(c),(d),(f).
As it is evident from the shape of v-t graph that acceleration of the particle is not uniform between time intervals t1 and t2. (since the given v-t graph is not straight). The equations (a), (b) and (e) represent uniform acceleration.