Question

A particle moves along the positive x-axis from origin with velocity, $V_0$ = 2m/s and a = 6(t - 1) where't' is in seconds. Which of these graphs are correct?

• A.
• B.
• C.
• D. None of these

Answer 1

Answer: (A), (C)

Here we are given a = 6(t-1)

Here, s = +x, v = +v

To find velocity - time relation using a = $\frac{dv}{dt}$ we have $\frac{dv}{dt}=6(t-1)$ ........ (1)

When velocity changes from $v_0$ to v during time t, integrating both sides with respect to time , we have

${\int }_{v_0}^{v}dv={\int }_0^{t}6(t-1)dt\Rightarrow v-v_0=3t^2-6t$

Substituting $v_0=2m/s$, we have ,

$v=\frac{dx}{dt}=3t^2-6t+2$

dx = $3t^2-6t+2)$ dt ..... (ii)

This gives x = $t^3-3t^2+2t$ = t (t -1) (t - 2).

The particle passes through the origin where x = 0 from equation (iii), we have t = 0, 1s and 2s.

That means, the particle crosses the origin twice at t =1s and t = 2s. Aftert = 2s, x is positive. Hence its

displacement points in positive x-direction and goes on increasing its magnitude.

The particle will come at rest when v= 0.

From (ii), we have t = $(1-\frac{1}{\sqrt{3}})s(=t_1,say)andt=(1+\frac{1}{\sqrt{3}})s=(=t_2,say).$

Hence the particle at t = $t_{1}and{t}_2.$ That signifies to and fro motion of the particle during first two seconds

as it starts retracing its path at time t = $t_{1}andt=t_2$.

The displacement (x), velocity (v) and acceleration (a) of the particle in different time intervals are given

in the followingtable and shown in the following graphs.