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 1.0Ns is applied to the block at time to t = 0 so that it starts moving along the x-axis with a velocity \[v(t)={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}\] where \[{{v}_{0}}\]is a constant and \[\tau \]= 4s. the displacement of the block at t= \[\tau \] will be?
Answer
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Hint: We should remember that distance is a scalar quantity and displacement is a vector quantity. Speed is the rate of change of distance and velocity is the rate of change of displacement. For vectors, we have to consider their directions.
Complete step by step answer:
Given velocity \[v(t)={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}\]
We know velocity is rate of change of displacement, so we can write the above equation as,
$
v(t)={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}} \\
\implies \dfrac{dx}{dt}={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}} \\
\implies dx={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}dt \\ $
Integrating both sides within the suitable limits, we get,
$
\int\limits_{0}^{x}{dx}=\int\limits_{0}^{\tau }{{{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}dt} \\
\implies x={{v}_{0}}\int\limits_{0}^{\tau }{{{e}^{\dfrac{-t}{\tau }}}dt} \\
\implies x={{v}_{0}}[\dfrac{{{e}^{\dfrac{-t}{\tau }}}}{\dfrac{-1}{\tau }}]_{0}^{\tau } \\
\implies x=-\tau {{v}_{0}}[{{e}^{-1}}-{{e}^{0}}] \\
\implies x=-4\times 25[0.37-1] \\
\therefore x=6.30 \\
$
So, the value of displacement rounds off to the nearest integer is 6 m.
Additional Information:
Because displacement is a vector quantity, we can add it algebraically. For vectors, we have to consider their directions. Here the vectors were perpendicular so the cosine component vanishes. Displacement and velocity are related as the velocity is the rate of change of displacement. If the displacement is zero then velocity will also be zero and if velocity is zero then there is no change in the displacement with respect to well defined initial point.
Note:
Here we have used definite integration because the body starts from rest and we were given the total time for which the body was in motion. If there were no such information given then we would have used indefinite integration which involves the use of constant of integration.
Complete step by step answer:
Given velocity \[v(t)={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}\]
We know velocity is rate of change of displacement, so we can write the above equation as,
$
v(t)={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}} \\
\implies \dfrac{dx}{dt}={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}} \\
\implies dx={{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}dt \\ $
Integrating both sides within the suitable limits, we get,
$
\int\limits_{0}^{x}{dx}=\int\limits_{0}^{\tau }{{{v}_{0}}{{e}^{\dfrac{-t}{\tau }}}dt} \\
\implies x={{v}_{0}}\int\limits_{0}^{\tau }{{{e}^{\dfrac{-t}{\tau }}}dt} \\
\implies x={{v}_{0}}[\dfrac{{{e}^{\dfrac{-t}{\tau }}}}{\dfrac{-1}{\tau }}]_{0}^{\tau } \\
\implies x=-\tau {{v}_{0}}[{{e}^{-1}}-{{e}^{0}}] \\
\implies x=-4\times 25[0.37-1] \\
\therefore x=6.30 \\
$
So, the value of displacement rounds off to the nearest integer is 6 m.
Additional Information:
Because displacement is a vector quantity, we can add it algebraically. For vectors, we have to consider their directions. Here the vectors were perpendicular so the cosine component vanishes. Displacement and velocity are related as the velocity is the rate of change of displacement. If the displacement is zero then velocity will also be zero and if velocity is zero then there is no change in the displacement with respect to well defined initial point.
Note:
Here we have used definite integration because the body starts from rest and we were given the total time for which the body was in motion. If there were no such information given then we would have used indefinite integration which involves the use of constant of integration.
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