Question

A particle is moving along the positive x-direction with an initial velocity of $12m{s}^{-1}$ is subjected to a retarding force that gives it a negative acceleration which varies linearly with time for the first $4$ seconds as shown and for the next $5$ seconds acceleration is constant.
Velocity of the particle at $t=4s.$

• A. $0m{s}^{-1}$
• B. $8m{s}^{-1}$
• C. $6m{s}^{-1}$
• D. $12m{s}^{-1}$

Answer 1

The correct option is C $6m{s}^{-1}$

From the graph:$a=\frac{-3t}{4}$ for $t<4s$

Given initial velocity $u=12m/s$

We know,

$\begin{array}{c}\frac{dv}{dt}=a\\ dv=adt\\ {\int }_u^{v}dv={\int }_0^4\frac{-3t}{4}dt\\ v-12{=}_0^4[\frac{-3t^2}{8}\\ v-12=-3.\frac{16}{8}=-6\\ v=12-6=6m/s\end{array}$

In positive $x$ direction.