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 graph correctly describes `(x_(1) - x_(2))` as function of time 't' ?

Options

A.
B.
C.
D.

Answer

B

Explanation

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.