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 $t=3\, s$ is

Answer 1

$\log v=\log 2+\left(\tan 45^{\circ}\right) \log t$
$(y=m x +c)$
$\Rightarrow \log v=\log 2+\log t=\log 2 t$
$\Rightarrow v = 2 t$
$\frac{d x}{d t}=2 t $
$\Rightarrow d x=2 t d t $
$\Rightarrow \int\limits_{0}^{x} d x=\int\limits_{0}^{3} 2 t d t$
$\Rightarrow x=\left(\frac{2 t^{2}}{2}\right)_{0}^{3}=9\, m$