A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m=0.4kg$ is at rest on this surface. An impulse of $1.0 N s$ is applied to the block at time $t=0$ so that it starts moving along the x-axis with a velocity $v\left(t\right)=v_0{e}^{-\frac{t}{\tau }}$ , where $v_0$ is a constant and $\tau =4$ . The displacement of the block, in meter, at $t=\tau$ is __________. Take $e^{-1}=0.37$

Solution

$v=v_0{e}^{-\frac{t}{\tau }}$
$v_0=\frac{J}{m}=2.5 m/s$
$v=v_0{e}^{-\frac{t}{\tau }}$
$\frac{dx}{dt}=v_0{e}^{-\frac{t}{\tau }}$
${\int }_0^{x}dx=v_0{\int }_0^{\tau }{e}^{-\frac{t}{\tau }}dt$
$\int e^{-x}dx=\frac{e^{-x}}{-1}$
$x=v_0{\left[\frac{e^{-\frac{t}{\tau }}}{-\frac{1}{\tau }}\right]}_0^{\tau }$
$x=2.5\left(-4\right)\left(e^{-1}-e^0\right)$
$x=25\left(-4\right)\left(0.37-1\right)$
$x=6.30$

Hence, the correct answer is 6.3.