Question

A particle is placed at the origin. It starts from rest at $t=0$ and undergoes an acceleration $a$ as shown in the figure. Which of the following options correctly represent velocity $v$ vs. time $t$ and displacement $x$ vs. time $t$ plots?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given motion has two phases
(i) accelerating phase from $t=0$ to $t=2s$ and
(ii) decelerating phase from $t=2s$ to $t=4s$.
In accelerating phase,
$a=3m/s^2($From $t=0s$ to $t=2s)$
$\Rightarrow v=3t(\because u=0)$
$\therefore$
Velocity will increase linearly with time
At $t=2s,v=6m/s$
Since $v=3t=\frac{dx}{dt}$
Integrating above equation, we get
$\begin{array}{}x=\frac{3}{2}{t}^2& \text{(1)}\end{array}$
Displacement varies parabolically with time.
In decelerating phase,
$a=-3m/s^2($From $t=2s$ to $t=4s)$
$\Rightarrow \int dv=\int -3dt$
$\Rightarrow v=-3t+c$
On putting $t=2s$ and $v=6m/s$ , we get
$c=12$
$\therefore v=-3t+12=\frac{dx}{dt}$
Integrating above equation,
$x=-\frac{3}{2}{t}^2+12t+C_1$
From equation (1), at $t=2s,x=6m$,
$C_1=-12$
Hence $x=-\frac{3}{2}{t}^2+12t-12$
Displacement varies parabolically with time.