Question

The figure shows the plot of velocity $(v)$ versus time $(t)$ on a log-log scale. Assuming straight line motion and the particle to start from origin, the distance (in metre) covered at the end of $t=3s$ is

Answer 1

Answer: 9

$\log v=\log 2+\left(\tan 45^{\circ }\right)\log t$
$(y=mx+c)$
$\Rightarrow \log v=\log 2+\log t=\log 2t$
$\Rightarrow v=2t$
$\frac{dx}{dt}=2t$
$\Rightarrow dx=2tdt$
$\Rightarrow \underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{3}{\int }}2tdt$
$\Rightarrow x={\left(\frac{2t^2}{2}\right)}_0^3=9m$