# Two body A and B are 10km apart such that B is in south of A. A and B start moving with same speed 20km/h eastward and northward respectively. Then find(a) Relative velocity of A w.r.t. B. (b) Minimum separation attained during motion. (c) Time lapse, from starting to attain minimum separationPlease give the answer with simle steps

Vibekjyoti Sahoo
145 Points
2 years ago

(1) Given,

A=> 20 km/hr E(eastward direction)
B=> 20Km/hr N (Northward).

velocity Va = 20
VB=20km/hr

velocity (r)= Va-Vb

Va-Vb= 20Km/hr( E)-20Km/hr (N)

In order to find "Va-Vb" with respect to B we consider it as positive by taking in opposite direction

Va- Vb= 20(E)+ 20(S)

To find relative velocity:-

|Va-Vb| => √ (20)²+(20)² ES (or) SE

|Va-Vb| => 20√2 SE

|Va-Vb| => 28.28 Km/hr SE.

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(2) Initial separation (ro)(distance)=10 Km

Relative velocity=28.28 Km/hr SE

Separation (r)= initial separation+ velocity

r=ro + 28.28 SE

At time t,
r= 10 Km+ 28.28 t SE

Since the direction has a common angle 45°

r= 10N+ 28.28t (Cos (45°)S+ sin(45°)E)

r= 10N+20 t E+ 20 t S

Since we has the initial separation at the north direction, so the south direction is reversed to the north.

r= 10N +20t E-20 tN
r= (10-20t)N +20 t E

We need to find the |r|

So we took |r|² for easier calculation.

|r|²= (10-20t)²N +(20t)²E
|r|² =100 - 400t + 400t^2 + 400t^2
|r|^2 = 800t^2 - 400t + 100
|r|^2 = 800(t^2 - t/2 + 1/8)
Multiply and divided by (2)
|r|^2 = 800[t^2 - 2(t)(1/4) + 1/16 + 1/8 - 1/16]

|r|^2 = 800[(t - 1/4)^2 + 1/16]

When the separation at |r|² it becomes minimum.

The minimum separation occurs at t= 1/4.

Sub . at |r|²,

|r|²= 800/16

|r| = √800/16

|r| = 5√2 Km .
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(3) time taken for minimum separation is t= 1/4

= 1/4×60 mins

= 15 mins.