Question

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 $1.0N$ s is applied to the block at time $t=0$ so that it starts moving along the x-axis with a velocity $v\left(t\right)=v_0{e}^{-t/\tau }$, where $v_0$ is a constant and $\tau$ = $4s$. The displacement of the block, in meters, at $t=0$ is __________.
Take $e^{-1}$ = $0.37$.

Answer 1

Answer: 6.3

$V=V_0{e}^{-t/\tau }$
$V_0=\frac{J}{m}=2.5\frac{m}{s}$
$V=V_0{e}^{-t/\tau }$
$\frac{dx}{dt}=V_0{e}^{-t/\tau }$
$\underset{0}{\overset{x}{\int }}dx=V_0\underset{0}{\overset{\tau }{\int }}{e}^{-t/\tau }dt$
$\therefore \int e^{-x}dx=\frac{e^{-x}}{-1}$
$x=V_0\left[\frac{e^{-t/\tau }}{-\frac{1}{\tau }}\right]$
$X=2.5(-4)\left(e^{-1}-e^0\right)$
$X=2.5(-4)(0.37-1)$
$X=6.30$