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A body is at rest at x =0 . At t = 0...

A body is at rest at ` x =0 ` . At ` t = 0`, it starts moving in the positive `x - direction` with a constant acceleration . At the same instant another body passes through ` x= 0 ` moving in the positive ` x - direction ` with a constant speed . The position of the first body is given by `x_(1)(t)` after time 't', and that of the second body by ` x_(2)(t)` after the same time interval . which of the following graphs correctly describes `(x_(1) - x_(2))` as a function of time 't' ?

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
B

For `1^(st) ` particle : " " It starts moving `(u_1=0 ) `which acceleration
` therefore " "x_1 =x_1 (t) =u_1t+(1)/(2) at^(2) =(1)/(2) at^(2) …….(1) `
For 2nd particle :
It is moving with constant velocity (v) ` therefore " " x_2 =x_2 (t) =vt " "...(ii)`
Relative position of particle 1 w.r.t. 2, ` " " x_1-x_2=x_1=(1)/(2) at^(2) -vt " "...(iii)`
Hence graph should be parabola.
Differentiating equation (iii) w.r.t time, we get the relative velocity of particle 1 w.r.t 2
` |v_(1,2) |(dx_(1.2))/(dt) =at -v, " " t=0,v_(1,2) =-v`
It means the slope of the parabola at t = 0 should be negative.
The parabola should open upwards. Hence graph ‘B’ fulfills the requirements.
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