neet exam preparation

2. M OT I O N I N A
STRAIGHT LINE


1. INTRODUCTION
Mechanics is a branch of Physics which deals with motion of bodies and the cause behind it. Motion of a body that
includes position and time can be determined with respect to other bodies. Branch of physics that deals with the
motion of particles and rigid bodies irrespective of the forces responsible for their motion is known as Kinematics.
When the size of a body is too small such that its motion can be described by a point mass moving along straight
line, motion is known as rectilinear motion or one-dimensional motion.

1.1 Motion
A body is said to be in motion when it changes its position with respect to the observer while it is said to be at rest
when there is no change in its position with respect to the observer. For instance, two passengers travelling in a
moving train are at rest with respect to each other but in motion for a ground observer.

1.2 Particle
Physically, a particle is considered as analogues to a point. A body with a definite size is considered as a particle
when all of its parts have same displacement, velocity and acceleration. The motion of any such body can be
studied by the motion of any point on that body.


1.3 Basic Definitions y

1.3.1 Position and Position Vectors rA A

Position of any point can be represented by its coordinates with respect to an
x
origin in Cartesian system. For example, point A can be represented by (xA, yA, zA). O
Figure 2.1
  ∧ ∧ ∧
rA = OA = x A i + y A j + z A k y

A B
Given the co-ordinates of two points A & B, the position vector of B w.r.t. A can be
rA
determined as follows:
rB
    x
AB =rB − rA ⇒ AB =( xB − x A ) ˆi + ( yB − y A ) ˆj + ( zB − z A ) kˆ
O
Figure 2.2

∧ ∧ ∧  
Illustration 1: Find the torque (τ) exerted by force i + j + 2k at point P (2, 3, 4) w.r.t origin. Given, τ =rxF

r : is the position vector of point at which force is acting w.r.t to the given point (in this case origin)  (JEE MAIN)

,2 . 2 | Motion in a Straight Line


Sol: The force F is given in Cartesian coordinates. Express the position vector of point P in Cartesian coordinates
 
and find the cross product τ =rxF
   
Here= r op= P [ P : Position vector of point P]
 ∧ ∧ ∧
⇒ r = 2 i + 3 j + 4k
 ∧ ∧ ∧ i j k
and F = i + j + 2k  ∧ ∧
  
∴ τ = r ×F 2 3 4 = i(2) − j(0) + k(
 −1) = τ = 2 i −k
∧ ∧ ∧ 1 1 2
= i + j + 2k


1.3.2. Distance and Displacement
y
A particle follows either a curve or a straight line when moving in the space. This curve
A
or line is known as its trajectory. Distance is the length of the path or trajectory covered
by the particle and displacement is the difference between the vectors of the first and rA rAB B
the last position on this path.
rB
Distance and displacement  are illustrated in the Fig. 2.3 where AB (curve length) is
 
the distance and vector ∆ rAB is the displacement ( rB − rA ) . Following points should be O x
considered about both the quantities: Figure 2.3
(a) Distance is a scalar quantity while displacement is vector.
(b) Displacement is always less than or equal to distance in magnitude.
(c) Displacement can be zero but distance cannot be zero for a moving body.


MASTERJEE CONCEPTS

For a particle moving in a straight line:
•• the distance travelled is always equal to the displacement when there is no change in direction,
i.e. distance travelled = |displacement|
•• Else, distance travelled is always greater than displacement,
i.e. distance travelled ≥ |displacement|
Vaibhav Gupta (JEE 2009 AIR 54)



Illustration 2: Find the distance and displacement of a particle travelling from one point to another, say from pt. A
to B, in a given path. (JEE MAIN)

50 m

10 m
40 m
A B



55 m

Figure 2.4

, P hysics | 2.3


Sol: Distance is length of the path travelled. Displacement is the vector
from initial point to final point.
Displacement
Total distance travelled = 10+50+40+55+ (40 - 10) = 185 m A B

Total Displacement = 50 + 55 =105 m
Deduction of displacement is from A (initial position) to B (final position) Figure 2.5


Illustration 3: If a particle travels a distance of 5 m in straight line and returns back to the initial point, then find
(i) Total distance travelled (ii) Total displacement (JEE ADVANCED)

Sol: Distance is length of the path travelled. Displacement is the vector from initial point to final point.
Total distance travelled = 5 + 5 = 10 m (since, initial and final 5m
position of the particle are same). A B
Displacement = 0 m (5-5=0, since the directions are opposite
and cancel with each other) Figure 2.6


1.3.3 Average Speed and Average Velocity
The total distance travelled by a particle divided by the total time taken is known as its average speed.
Total distance travelled
Average speed =
Total time taken
displacement rB − rA ∆rAB
While the average velocity is defined as v av = = or v =
Or
time elapsed tB − t A ∆t
Both average speed and average velocity are expressed in ms-1 or kmh-1, the former is a scalar quantity while the
latter is a vector.


MASTERJEE CONCEPTS

For a moving body:
•• Average velocity can be zero but average speed cannot be zero.
•• The magnitude of average velocity is always less than or equal to the average speed because,
displacement ≤ dis tance

displacement distance
≤
time elapsed time elapsed
therefore,
•• Average speed does not mean the magnitude of the average velocity vector.
Vaibhav Krishnan (JEE 2009 AIR 22)



Illustration 4: If a train moves from station A to B with a constant speed v =40 km/h and returns back to the initial
point A with a constant speed V2=30 km/h, then calculate the average speed and average velocity.  (JEE MAIN)

Sol: Average speed is distance covered divided by time taken. Distance is length of the path travelled. Average
velocity is displacement divided by time taken. Displacement is the vector from initial point to final point.
s
Let the distance AB = s, Time taken by train from A to B, t1 =
v1