# RELATIVE MOTION

The motion between two moving bodies is known as relative motion.

The motion of a moving body with respect to another moving body, is known as the relative motion of the first body with respect to second body.

5.5.1. Relative Motion between Two Bodies Moving in Straight Lines

Fig. 5.6 (a) shows the two bodies A and B which are moving along parallel lines in the same direction.

Fig. 5.6

Let vA = Absolute velocity of A

vB = Absolute velocity of B

Let vA > vB To find the relative velocity of A with respect to B, make the body B stationary. This is possible if a velocity equal to vB is applied on the body B in the opposite direction as shown in Fig. 5.6 (b). Then we shall have to apply a
velocity equal to vB on the body. A also in the opposite direction. The resultant of these two velocities acting on A will represent the relative
velocity of A with respect to B. As vA > vB, hence the resultant velocity will be in the direction of vA and equal to (vA – vB). This resultant velocity is the relative velocity of A with respect to B.

The relative velocity of A with respect to B in vector form is also given by,

vAB = Vector difference of vA

and uB

Graphically, the relative velocity of A with respect to B can also be determined as [Refer to Fig. 5.6 (e)] given below:

(i) Take any point o. From o, draw oa parallel to velocity vA and take oa = vA to some
suitable scale.

(ii) From o, also draw ob parallel to velocity vB Take ob = vB Joint b to a .

Then ba represents the relative velocity of A with respect to body B in magnitude and direction. Hence relative velocity of A with respect to B may also be written in the vector form. [Refer to Fig. 5.6 (e)].

5.5.2. Relative Motion between Two Bodies Moving Along Inclined Lines

Fig. 5.7 (a) shows the two bodies A and B moving

along inclined lines with absolute velocities vA and vB respectively. To find the relative velocity of A with respect to E, make the body B stationary. This is possible if a velocity equal to vB is applied on the body B in the opposite direction as shown in Fig. 5.7 (b). Then we shall have to apply a velocity equal to vB on the body A also in opposite direction.

Fig. 5.7

The resultant gives the relative velocity of A with respect to B as shown in Fig. 5.7 (b).

The vector vA in Fig. 5.7 (c) represents the velocity vA in magnitude and direction, whereas the vector ob represents the velocity, in magnitude, and direction. Join points b and a. Then relative velocity of A with respect to B is written as vAB and is given by

Problem 5.25. Two roads which are at an angle of 600, intersect at a point 0. A car 1 is moving with a velocity of 72 km/hr away from point a along x-axis at an instant of time. At the same time another car 2 is moving away from point a with a velocity of
48 km/hr along the road 2.

Determine the velocity of car 1 with respect to car 2 .

Fig. 5.8 (a)

As tanΘ is negative, hence e should be in second or fourth quadrant. If e is in second
quadrant, then x-component should be – ve and y-component should be positive. But here y-component is negative, this means Θ lies in fourth quadrant.

Θ = tan1 (- 0.865) = 40.86° or (360 – 40.86) = 319.14°. Ans.

Alternate method (Graphical Method)

(i)                           Take any point o. From o draw a line oa parallel to the
velocity of car 1 or V­1 in magnitude and direction. Cut oa = 20 m/s.

(ii)                        From o, also draw line ob parallel to the velocity of car 2
or V2 in magnitude and direction. Cut ob = 13.33 m/s. Join b to a.
Then vector ba is the relative velocity of car 1 with respect to
car 2.

By measurement vector ba = 17.6 m/s and angle Θ = 40.86°.

θ= 360 – 40.86 = 319.14°. Ans.

Note. For graphical method both the velocities should be directed either away from the origin (point 0) or towards the origin.

Problem 5.26. If in the previous problem, the car 1 is at a distance of 100 m from a along x-axis and is moving with a speed of 72 km/hr towards a at the time of observation and car 2 is mooing with a speed of 48 11m I hr. along road 2 and away from the origin as shown in Fig. 5.9, then find the velocity of car 1 with respect to car 2.

Sol. Given:

Distance of car 1 from 0 along .r-axis = 100 m

The car 1 is moving along x-axis and towards origin.

Hence the velocity of car 1 in vector form is written as

V1 = – 20i   [- ve sign is due to opposite direction of car 1]

The velocity of car 2 in vector form is written as

V2 = 13.33 [cos 60i + sin 60j] = 6.66i + 11.54j

The velocity of car 1 with respect to car 2 is given by,

V12 = V1 – V2 = – 20i – [6.66i + 11.54j]

= – 26.66i – 11.54j. Ans.

Graphical Method          Fig. 5.9 (a)

(i)                           Take any point o From o draw a line oa parallel to velocity VI in magnitude and direction. Take oa = 20 m/s,

(ii)                        From o, also draw a line ob parallel to velocity V2 in magnitude and direction. Take ob = 13.33 m/s. Now vector ba represents the velocity of 1 with respect to 2 in magnitude and direction. Measure vector ba.

Problem 5.27. At an instant of time, a car 1 at a distance of 120 m from point o along
x-axis moving at a velocity of
72 km/hr decelerates at 1. 5 m/s2 as it approaches the point o. At the same instant, a car 2, is moving along a road which is at an angle of 600 with the road on which car 1 is moving. This car is moving away from o with a velocity of 48 km / hr and accelerates at 2 m/s2 as shown in Fig. 5.10. Find the velocity of car 1 with respect to car 2 after three seconds.

= – 20i m/s (in vector notation)

(- ve sign is due to car 1 is moving towards 0 along x-axis)

Deceleration = 1.5 m/s2 (towards o)

or Acceleration,  a = 1.5m/s2 (away from o along x-axis)

= 1.5i m/s2 (in vector notation)

(+ ve sign is as the acceleration is acting along z-axis away from o)

Time, t = 3 seconds

Let v1 = Final velocity of car 1 after 3 seconds

Then v1 = u1 + at = (- 20i) + (1.5i) × 3

(u1 = – 20i ; a1 = 1.5i)

= – 20i + 4.5i = – 15.5i

V1 = -15.5i

[Final velocity of car 1, in vector notation]

This velocity is at an angle of 60° with x-axis. Hence in vector notation, it is given by

u2 = 13.33 (cos 600)i + 13.33 (sin 600)j

= 6.67i + 11.54j

Acceleration, a = 2 m/s2. This acceleration is at an. angle of 60° with x-axis. Hence this acceleration in vector form is given by,

a = 2(cos 600)i + 2 sin (600)j = 1.0i + l.732j

Time t = 3s

Let v2 = Final velocity of car 2 after 3 seconds.

v2 = u2 + at

= (6.67i = 11.54j) + (1.0i + 1.732j) × 3

= 6.67i + 11.54j + 3i + 5.196j

= 9.67i + 16.736j

V2 = 9.67i + 16.736j

Now the velocity of car 1 with respect to the velocity of car 2 after 3 seconds

V12 = V1 – V2 =-15.5i – [9.67i + 1.6.736j]

= – 25.17i- 16.736j m/s. Ans.