Question

Figure shows the acceleration-time graph of a particle moving along a straight line. After what time the particle acquires its initial velocity?

Answer 1

From the graph, slope of oblique line $=\frac{2-0}{1-2}=-2$ for $t>1$

$\therefore$ from similar triangles properties between upper and lower triangle $\frac{a}{t-2}=\frac{2}{2-1}$ $⟹$ Magnitude of acceleration $|a\left(t\right)|=2t-4$ for $t>2$ s

Magnitude of acceleration $|a\left(t\right)|=2t-4$ for $t>2$ s

Area under the a-t graph gives the change in velocity of the particle during that period.

$\therefore$ $\Delta V=area$ $of$ $rectangular$ $portion$+ $area$ $of$ $upper$ $trianle$-$area$ $of$ $lower$ $triangle$

$\therefore$ $\Delta V=2\times 1+\frac{1}{2}\times 1\times 2-\frac{1}{2}\times \left(a\right(t\left)\right)\times (t-2)$

$\therefore$ $\Delta V=2\times 1+\frac{1}{2}\times 1\times 2-\frac{1}{2}\times (2t-4)\times (t-2)$

$⟹$ $\Delta V=3-(t-2{)}^2$

Let the particle acquires its initial velocity again after time $t_1$ i.e

change in velocity=$0$

$\Delta V=0$ at $t=t_1$

$\therefore$ $0=3-(t_1-2{)}^2$ $⟹t_1=2+\sqrt{3}$ s and $t_1=2-\sqrt{3}$ s (Not possible)(Because for $t_1=2-\sqrt{3}$ s Acceleration is always positive so change in velocity can not be zero)

Thus the particle acquires its initial velocity again at $t=2+\sqrt{3}$ s