Question

Acceleration time graph of a particle is shown. Work done by all the forces acting on the particle of mass m in time interval $t_{1}$ and $t_{2}$ while $a_{1}$, is the acceleration at time $t_{1}$, is given by : (at $t = 0,$ particle was at rest)

• A. $\frac{ma^{2}_{1}}{4t_{1}}\left(t^{3}_{2}-t^{3}_{1}\right)$
• B. $\frac{ma^{2}_{1}}{8t^{2}_{1}}\left(t^{4}_{2}-t^{4}_{1}\right)$
• C. $\frac{ma^{2}_{1}}{4t^{2}_{1}}\left(t^{4}_{2}-t^{4}_{1}\right)$
• D. $\frac{ma_{1}}{2t_{1}}\left(t^{2}_{2}-t^{2}_{1}\right)$

Answer 1

Answer: (B)

$W=K_{f}-K_{i}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)$
Area under acceleration-time graph is change in velocity.
So, $W =\frac{1}{2} m\left[\left(\frac{1}{2} a_{2} t_{2}\right)^{2}-\left(\frac{1}{2} a_{1} t_{1}\right)^{2}\right]$
$=\frac{1}{8} m\left[\frac{t_{2}^{4}}{t_{1}^{2}} a_{1}^{2}-a_{1}^{2} t_{1}^{2}\right]$
(since, $a_{2}=\frac{t_{2}}{t_{1}} a_{1}$)
$=\frac{m a_{1}^{2}}{8 t_{1}^{2}}\left(t_{2}^{4}-t_{1}^{4}\right)$