Question

A particle starts from rest with a time-varying acceleration, $a=(2t-4)$. Here $t$ is in $s$ and $a$ in $m/s^2$. After what time (in $s$) particle comes to rest?

• A. $1$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is B $4$

Acceleration – time graph of the particle is a straight line as shown in the figure. Area enclosed by the $a-t$ graph and the $x$-axis gives the change in velocity .

Particle will come to rest, when area above time axis and below becomes equal.
Area below time axis, $A_1=\frac{1}{2}\times 4\times 2=4$
Area above time axis, $A_2=\frac{1}{2}\times t\times a=A_1=4$
$\begin{array}{}\Rightarrow at=8& \text{(1)}\end{array}$
Slope of the line, $m=\frac{4}{2}=2$
Hence $\begin{array}{}\frac{a}{t}=2& \text{(2)}\end{array}$
Solving (1) and (2), we get
$t=2s$
Hence after $t+2$ i.e. $4\text{sec}$, particle comes to rest.

Alternate:
Given, particle starts from rest i.e. at $t=0,v=0$
As we know that
$a=\frac{dv}{dt}\Rightarrow v\left(t\right)={\int }_0^{t}adt$
$\Rightarrow v\left(t\right)={\int }_0^t(2t-4)dt$
$\Rightarrow v\left(t\right)=t^2-4t$
The particle will come to rest only if $v=0$
$\Rightarrow t^2-4t=0\Rightarrow t(t-4)=0$
$\Rightarrow t=0,4$
Hence, after $4\text{sec}$, particle will come to rest.