Question

Starting from rest a particle moves in a straight line with acceleration $a=\{2+|t-2|{\text{m/s}}^2\}$. What will be the velocity of particle in m/s at the end of $4\text{s}$?

Answer 1

Acceleration can be written as
$a=2+2-t$ or $a=4-t$ for $t\le 2\text{s}$
and $a=2+t-2$ or $a=t$ for $t\ge 2\text{s}$
Therefore, acceleration-time graph of the particle will be as in the figure.
Now, since
$\int dv=\int adt\Rightarrow v_f-v_i=\text{area under}a-t\text{graph}$
or $v_f-0=(4\times 4)-\frac{1}{2}\left(4\right)\left(2\right)=12\text{m/s}$
or velocity of particle at the end of $4\text{s}$ is $12\text{m/s}$.