A particle moves along a straight line with a variable acceleration given in the acceleration-displacement $\left(\mathrm{a}-\mathrm{S}\right)$ curve as shown in the figure. Determine the velocity (in $\mathrm{m} {\mathrm{s}}^{-1}$) of the particle after it has travelled a distance of $30 \mathrm{m}$. The initial velocity is $10 \text{m }{\mathrm{s}}^{-1}$.
 

Solution

Area under curve is $=\frac{1}{2}\times 10\times 30$
$=150$ ……..(i)
Area under the curve is also equal to $=\frac{v^2-u^2}{2}$ …….(ii)
From (i) and (ii)
$\frac{1}{2}\left(v^2-u^2\right)=150$
$\left(v^2-u^2\right)=300$
$v^2=u^2+300$
$v^2={\left(10\right)}^2+300$
$v=\sqrt{400}=20 \text{m }{\mathrm{s}}^{-1}$

Hence, the correct answer is 20.