Question

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m=0.4$kg is at rest on this surface. An impulse of $1.0$N s is applied to the block at time to $t=0$ 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 $\tau =4$s. The displacement of the block, in metres, at $t=\tau$ is ___________.
[Take $e^{-1}=0.37$]
Round off your answer to nearest integer.

Answer 1

$v=v_o{e}^{\frac{-t}{\tau }}$

$v_o=\frac{J}{m}=2.5m/s$

$v=v_o{e}^{\frac{-t}{\tau }}$

$\frac{dx}{dt}=v_o{e}^{\frac{-t}{t}}$

${\int }_0^{x}dx=v_o{\int }_0^t{e}^{\frac{-t}{\tau }}dt$

$\int e^{-x}dx=\frac{e^{-x}}{-1}$

$x=v_o{\left[\frac{e^{\frac{-t}{\tau }}}{\frac{1}{\tau }}\right]}_0^t$

$x=2.5(-4)(e^{-1}–e^0)$

$x=25(-4)(0.37-1)$

$x=6.30\approx 6$