Question

A particle starting from rest undergoes acceleration given by $a=|t–2|{\text{m/s}}^2$ where, $t$ is time in second. Velocity of particle after $4\text{sec}$ is

• A. $1\text{m/s}$
• B. $2\text{m/s}$
• C. $8\text{m/s}$
• D. $4\text{m/s}$

Answer 1

The correct option is D $4\text{m/s}$

As we know that,
$\frac{dv}{dt}=a$
$\Rightarrow \int dv=\int adt$
Since, $a=(t-2),$ has negative value for $t<2$ and positive value for $t>2$, so, the value of function $a=|t-2|$ will be
$a=-(t-2);t<2$
and $a=(t-2);t>2$
$\Rightarrow {\int }_0^{v}vdv={\int }_0^2-(t-2)dt+{\int }_2^4(t-2)dt$
$\Rightarrow v={[\frac{-t^2}{2}+2t]}_0^2+{[\frac{t^2}{2}-2t]}_2^4$
$\Rightarrow v=2+2=4\text{m/s}$