# Motion as a Straight Line NCERT Textbook PDF

NCERT Solutions for Class 11 Physics Chapter 3‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Physics Chapter 3 Exercise Solution’ using the download button.

### Chapter 3: Motion as a Straight Line

#### 3.1 Introduction

Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves into and out of our lungs, and blood flows in arteries and veins.

We see leaves falling from trees and water flowing down a dam. Automobiles and planes carry people from one place to the other.

The earth rotates once every twenty-four hours and revolves round the sun once a year.

The sun itself is in motion in the Milky Way, which is again moving within its local group of galaxies.

Motion is a change in the position of an object with time. How does the position change with time?

In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity and acceleration.

We shall confine ourselves to the study of the motion of objects along a straight line, also known as rectilinear motion.

For the case of rectilinear motion with uniform acceleration, a set of simple equations can be obtained.

Finally, to understand the relative nature of motion, we introduce the concept of relative velocity.
In our discussions, we shall treat the objects in motion as point objects.

This approximation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time.

In a good number of situations in real-life, the size of objects can be neglected and they can be considered as point-like objects without much error.

In Kinematics, we study ways to describe motion without going into the causes of motion.

What causes motion described in this chapter and the next chapter forms the subject matter of Chapter 5.

#### 3.2 Position, Path Length And Displacement

Earlier you learned that motion is a change in the position of an object with time. In order to specify the position, we need to use a reference point and a set of axes.

It is convenient to choose a rectangular coordinate system consisting of three mutually perpendicular axes, labeled X-, Y-, and Z- axes.

The point of intersection of these three axes are called origin (O) and serve as the reference point.

The coordinates (x, y. z) of an object describe the position of the object with respect to this coordinate system.

To measure time, we position a clock in this system. This coordinate system along with a clock constitutes a frame of reference.

If one or more coordinates of an object change with time, we say that the object is in motion.
Otherwise, the object is said to be at rest with respect to this frame of reference.

The choice of a set of axes in a frame of reference depends upon the situation. For example, for describing motion in one dimension, we need only one axis. To describe motion in two/three dimensions, we need a set of two/ three axes.

### NCERT Solutions Class 11 Physics Chapter 3 Motion as a Straight Line

Q.5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?Ans:Speed of the jet airplane, VA= 500 km/hSpeed at which the combustion products are ejected relative to the jet plane, VB – VA= – 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 on the ground, VB – 500 = – 1500VB = – 1500 + 500 = – 1000 km/h

Q6 A car moving along a straight highway with a speed of 126 km h–1 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.

The initial velocity of the car = u

Final velocity of the car = v

Distance covered by the car before coming to rest = 200 m

Using the equation,

v = u + at

t = (v – u)/a = 11.44 sec.

Therefore, it takes 11.44 sec for the car to stop.

Q.7. Two trains A and B of length 400 m each are moving on two parallel tracks with a
uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of
B decides to overtake A and accelerates by 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:

Length of the train A and B = 400 m

Speed of both the trains = 72 km/h = 72 x (5/18) = 20m/s

Using the relation, s = ut + (1/2)at2

Distance covered by the train B

SB = uBt + (1/2)at2

Acceleration, a = 1 m/s

Time = 50 s

SB = (20 x 50) + (1/2) x 1 x (50)2

= 2250 m

Distance covered by the train A

SA = uAt + (1/2)at2

Acceleration, a = 0

SA = uAt  = 20 x 50 = 1000 m

Therefore, the original distance between the two trains = SB – SA = 2250 – 1000 = 1250 m

Q. 8. On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars B and
C approach car A in opposite directions with a speed of 54 km/h 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?

Ans:

The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s

Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s

Speed of car C = – 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite direction)

Relative speed of A w.r.t C, VAC= VA – VB = 10 – (-15) = 25 m/s

Relative speed of B w.r.t A, VBA = VB – VA = 15 – 10 = 5 m/s

Distance between AB = Distance between AC = 1 km = 1000 m

Time taken by the car C to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 s

If a is the acceleration, then

s = ut + (1/2) at2

1000 = (5 x 40) + (1/2) a (40) 2

a = (1000 – 200)/ 800 = 1 m/s2

Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s2
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