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 v_{A} = Absolute velocity of A

v_{B} = Absolute velocity of B

Let v_{A} > v_{B} To find the relative velocity of A with respect to B, make the body B stationary. This is possible if a velocity equal to v_{B} 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 v_{B} 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 v_{A} > v_{B}, hence the resultant velocity will be in the direction of v_{A} and equal to (v_{A} – v_{B}). 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,

v_{AB} = Vector difference of v_{A}

and u_{B}

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 v_{A} and take oa = v_{A} to some

suitable scale.

(ii) From o, also draw ob parallel to velocity v_{B} Take ob = v_{B} 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 v_{A} and v_{B} 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 v_{B} 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 v_{B} 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 v_{A} in Fig. 5.7 (c) represents the velocity v_{A} 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 v_{AB} and is given by

**Problem 5.25**. *Two roads which are at an angle of 60 ^{0}, 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.

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 = tan^{–}^{1} (- 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 V_{1} in magnitude and direction. Cut oa = 20 m/s.

(ii) From o, also draw line ob parallel to the velocity of car 2

or V_{2} 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

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

The velocity of car 2 in vector form is written as

V_{2} = 13.33 [cos 60i + sin 60j] = 6.66i + 11.54j

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

V_{12} = V_{1} – V_{2} = – 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 V_{2} 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*

*/s*

^{2}

*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*

*/s*

^{2}

*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/s^{2} (towards o)

or Acceleration, a = 1.5m/s^{2} (away from o along x-axis)

= 1.5i m/s^{2} (in vector notation)

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

Time, t = 3 seconds

Let v_{1} = Final velocity of car 1 after 3 seconds

Then v_{1} = u_{1} + at = (- 20i) + (1.5i) × 3

(u_{1} = – 20i ; a_{1} = 1.5i)

= – 20i + 4.5i = – 15.5i

V_{1} = -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

u_{2} = 13.33 (cos 60^{0})i + 13.33 (sin 60^{0})j

= 6.67i + 11.54j

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

a = 2(cos 60^{0})i + 2 sin (60^{0})j = 1.0i + l.732j

Time t = 3s

Let v_{2} = Final velocity of car 2 after 3 seconds.

v_{2} = u_{2} + at

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

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

= 9.67i + 16.736j

V_{2} = 9.67i + 16.736j

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

V_{12} = V_{1} – V_{2} =-15.5i – [9.67i + 1.6.736j]

**= – ****25.17i- 16.736j m/s. Ans. **

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