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  4. 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 t = 0 so that it starts moving along the x-axis with a velocity v(t) = v0e-t/τ, where v0 is a constant and τ = 4 s. The displacement of the block, in meters, at t = 0 is . Take e-1 = 0.37.

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Q. 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 $t = 0$ 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 $\tau$ = $4\, s$. The displacement of the block, in meters, at $t = 0$ is __________.
Take $e^{-1}$ = $0.37$.

JEE AdvancedJEE Advanced 2018

Solution:

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