Question

The figure shows the v - t graph of a particle moving in straight line. Find the time (in $s$) when particle returns to the starting point.

Answer 1

The particle returns to its starting point, when the net displacement of the particle is zero.

As the net displacement of the particle is given by area under $v-t$ graph, we can say:

Area under $v-t$ graph $=0$
Consider $t$ as the time when area becomes zero, then
$\frac{1}{2}\times 25\times 20-\frac{1}{2}\times (t-25)\times \left(v\right)=0$
Here, $v$ is the velocity of particle at time $t$.
Now we need to calculate $v$ in terms of $t$.
For that, we can equate slopes of both adjacent triangles:
$\frac{20}{5}=\frac{v}{t-25}$
$\Rightarrow v=4(t-25)$

$\therefore 250-\frac{1}{2}\times (t-25)\times 4(t-25)=0$
$250-2(t-25{)}^2=0$
$(t-25{)}^2=125$
$t-25=\pm 5\sqrt{5}$
$t=25\pm 5\sqrt{5}$

As $t>25$, we have to take $t=25+5\sqrt{5}$
$\therefore t=25+5\times 2.2=36s$