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' ?

Options

A.
B. .
C. .
D. .

Answer

B

Explanation

For `1^(st)` particle :
It starts moving `(u_1 = 0)` with constant acceleration
`x_1 = x_1(t) = u_1 t + (1)/(2) at^2 = (1)/(2) at^2`…(i)
For `2^(nd)` particle :
It is moving with constant velocity (v)
`x_2 = x_2(t) = vt` …(ii)
Relation position of particle '1' w.r.t. '2'
Hence `x_1 - x_2 = x_(1,2) = (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`
or `|v_(1,2)| = (dx_(1,2))/(dt) = at - v`
As particle '1' starts moving from rest, hence at `t = 0`
`v_(1,2)` should be negative.
It means the slope of the parabola at `t = 0` should be negative. The parabola should open up. Hence graph 'b' fulfil the requirement.