NCERT Solution (Class 11) > Physics Chapter 3 Motion In A Straight Line

NCERT Solution (Class 11) > Physics Chapter 3 Motion In A Straight Line

NCERT Solution Class 11 Physics Chapter 3 Motion In A Straight Line: National Council of Educational Research and Training Class 11 Physics Chapter 3 Solution – Motion In A Straight Line. Free PDF Download facility available at our website.

Board

NCERT
Class

11

Subject

Physics
Chapter

3

Chapter Name

Motion In A Straight Line
Topic

Exercise Solution

Motion In A Straight Line Chapter all Questions and Numericals Solution

 

3.1) 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.

(b) a monkey sitting on top of a man cycling smoothly on a circular track.

(c) a spinning cricket ball that turns sharply on hitting the ground.

(d) a tumbling beaker that has slipped off the edge of a table.

ANSWER-

a) Here both a and b are considered point objects.

The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be considered as a point sized object.

b) The monkey can be considered as a point object in this case as The size of a monkey is very small as compared to the size of a circular track.

c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply. Hence, the cricket ball cannot be considered as a point object.

d) The size of a beaker is comparable to the height of the table Hence it can not be considered as a point

 

3.2) The position-time (x-t) graphs for two children A and B returning from their school to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below ;

(a) (A/B) lives closer to the school than (B/A)

(b) (A/B) starts from the school earlier than (B/A)

(c) (A/B) walks faster than (B/A)

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

(e) (A/B) overtakes (B/A) on the road (once/twice).object

ANSWER-

(a) A lives closer to school than B as distance OP is smaller than OQ on y axis.

(b) A starts from school earlier than B.

(c) B walks faster than A as slope of the x-t graph shows velocity. As slope of line B looks more than slope of A hence B walks faster.

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

(e) from graph it is clear that B overtakes A once on the road.

 

3.3)

3.9)

3.10) 

 

3.11) 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

(b) with zero speed may have non-zero velocity,

(c) with constant speed must have zero acceleration,

(d) with positive value of acceleration must be speeding up.

ANSWER-

(a)True. Acceleration is result of change in velocity so at any instant as long as change in velocity is nonzero acceleration is present.

(b) False. Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity is ultimately zero.

(c) True. Acceleration is change in velocity with respect to time. If velocity is constant which means change in velocity is zero which gives acceleration zero.

(d) False.

 

3.12)

 

3.13) 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;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].

ANSWER-

(a) The magnitude of displacement over an interval of time is the shortest distance  between the initial and final positions of the particle whereas The total length of  path a particle is the actual path covered by the particle in a given interval of time.

For example, suppose a particle moves from point A to point B and then come again to point A then shortest distance is zero between initial and final position while total path length is AB+BA = 2AB.

(b)average velocity over an interval of time is defined as total displacement per unit time while Average speed of a particle over an interval of time is defined as the total path length divided by the time interval. As total path length and displacements are different quantities hence average velocity and Average speed both are different.

 

3.14)

 

In case you are missed :- Previous Chapter Solution

 

 

3.15) In Exercises 3.13 and 3.14, 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?

ANSWER-

Instantaneous velocity is given by rate of change of displacement with respect to time where time tends to zero. Which means it is defined for very short interval of time. During this small interval of time we can assume that particle can’t change its direction hence total displacement and total path length are equal hence both instantaneous speed and velocities are also equal.

 

3.16) Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

ANSWER-

  • from graph a at given time we are getting two values of x which shows particle is present at two different position at same time which is not possible in one dimension motion.
  • from graph a at given time we are getting two values of velocities v hence it does no represents one dimensional motion.
  • Graph of speed vs time is given in which speed is negative for some value of time t. hence it is not possible to have negative value of speed as it is scalar quantity.
  • This graph is not one-dimensional motion of a particle as path length cannot decrease with time.

 

3.17) Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line fort < 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

ANSWER-

It is not correct to say so as   for t < 0 we have  x = 0 which means particle does not move at all. For physical context we can consider the stone hold by someone and then leave to fall.

 

3.18)

 

3.19 Suggest a suitable physical situation for each of the following graphs (Fig 3.22):

ANSWER-

a) The given x-t graph shows that initially a body was at rest from initial part of graph after that slope of graph is positive which shows increase in velocity up to point A. Then, its velocity increases with time in the opposite direction and becomes zero.

 

b) In the given v- t graph, magnitude of velocity decreases with a time. A similar situation is when a rubber ball is dropped on the floor from a height. It strikes the floor with some velocity and rebounds with velocity less than original velocity and finally becomes zero.

 

c) From a-t graph as acceleration is zero in graph which indicates that the body is moving with a uniform velocity. After that sudden rise in acceleration is observed and within short time span it drops to zero means the body again starts moving with the same constant velocity.

 

3.20) 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, – 1.2s.

ANSWER-

At t = 0.3 s position is negative. As graph shows negative slope hence velocity is negative but acceleration in simple harmonic motion is given by,

a = – ω2x where x is displacement from mean position hence acceleration is negative.

At t = 1.2 s. between t= 1 to 1.5 sec. x is increasing hence positive. Slope is positive hence velocity is positive and acceleration is negative as x is positive .

At t = -1.2s x is negative. Both x and t are negative then the velocity comes to be positive. From acceleration formula a = – ω2x we can say that acceleration will be positive.

 

3.21) Figure 3.24 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.

ANSWER-

The average speed of a particle shown in the graph in a particular interval of time. It is clear from the graph that the slope is maximum in intervals 3 hence the average speed is maximum in intervals 3and slope is minimum in interval 2 hence the average speed of the particle is least in interval 2. The sign of average velocity is depends upon slope of x-t graph. Positive the slope positive will be the sign and vice the versa. Here for intervals 1 and 2 the slope is positive hence the velocity is also positive while for interval 3 slope is negative hence the velocity is also negative.

 

3.22) Figure 3.25 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?

ANSWER-

We know that Acceleration is given by the slope of the speed-time graph. By looking at graph the slope of the given speed-time graph is maximum in interval 2 hence average acceleration will be the greatest in interval 2. Average speed of the particle is given by y axis so from graph. It is clear that interval 3 has greatest average speed.

 

For interval 1: The slope of graph is positive. Hence, acceleration is positive as slope represents the acceleration. Similarly, the speed of the particle is positive in this interval. As it is in first quadrant of graph.

 

For interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative as slope represents the acceleration and speed is positive as it is in first quadrant of graph.

 

For interval 3: graph looks horizontal hence the slope of the speed-time graph is zero. Hence, acceleration is also zero and speed is positive as it is in first quadrant of graph.

 

Additional Exercise

3.23)

 

3.25) On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h–1 (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 h–1. 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 ?. (b) speed of the child running opposite to the direction of motion of the belt ? (c) time taken by the child in (a) and (b) ? Which of the answers alter if motion is viewed by one of the parents ?

ANSWER-

Speed of the belt, vB = 4 km/h

Speed of the boy, vc= 9 km/h

(a)in this case Since direction of motion of boy and direction of motion of the belt is same hence his speed

Relative to the observer is calculated by addition of speeds as belt speed will help boy

vBC= vC + vB = 9 + 4 = 13 km/h

(b) in this case Since direction of motion of boy and direction of motion of the belt is opposite hence his speed Relative to the observer is calculated by subtraction of speeds as belt speed will oppose the boy vCB = vC+ – vB = 9 – 4 = 5 km/h

(C) to find the time taken we can say that speed of the child in both the cases is same for the parents as both parents are on the same belt which shows that if the observer is one of the parents then the results obtained in (a) and (b) will change and velocity of child will be v = 9 km/h and distance between the child’s parents is 50 m = 0.05 km hence

The time taken by the child is,

t = D/V = 0.05/9 x 3600 sec = 20 sec.

 

3.26)

 

3.27)

 

3.28) 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 t1to t2:

(a) x(t2 ) = x(t1) + v (t1) (t2 – t1) +(½) a (t2 – t1)2

(b) v(t2 ) = v(t1) + a (t2 – t1)

(c) vaverage = (x(t2) – x(t1))/(t2 – t1)

(d) aaverage = (v(t2) – v(t1))/(t2 – t1)

(e) x(t2 ) = x(t1) + vaverage (t2 – t1) + (½) aaverage (t2 – t1)2

(f) x(t2 ) – x(t1) = area under the v-t curve bounded by the t-axis and the dotted line shown.

ANSWER-

Given velocity time graph is non-uniform in nature hence average velocity average acceleration and area under the v-t curve is used hence formula c, d, and f can be used.

In case you are missed :- Next Chapter Solution

Updated: April 17, 2023 — 11:21 am

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