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The particle that move along a straight line has velocity v(t)=t^2e^{-3t}m\,s' after t seconds....

Question:

The particle that move along a straight line has velocity {eq}v(t)=t^2e^{-3t}m\,s' {/eq} after {eq}t {/eq} seconds. Determine the distance {eq}x(t) {/eq} that it will travel during the first {eq}t {/eq} seconds.

Time, Distance & Velocity:

The time by which the particle can travel is known as the distance and the rate of change of distance will give speed. This if speed is given, then integrating that with tine will distance traveled.

Answer and Explanation:


In the problem, we have been given the particle that moves along a straight line has velocity {eq}v(t)=t^2e^{-3t}m\,s' {/eq} after {eq}t {/eq} seconds. We need to determine the distance {eq}x(t) {/eq} that it will travel during the first {eq}t {/eq} seconds.

So this is done by knowing the fact that, the distance(S) is given as:

{eq}S=\int_{0}^{t}t^2e^{-3t} dt\\ {/eq}

So solving this definite integral, we get the total distance in t seconds.

{eq}S=\int_{0}^{t}t^2e^{-3t} dt\\ {/eq}

Applying the substitution method,

{eq}u=-3t\\ \rightarrow \:du=-3dt\\ {/eq}

we get, the substitute integral as:

{eq}\int _0^{-3t}-\frac{e^uu^2}{27}du\\ {/eq}

Now applying the integration by parts, we have:

{eq}w=u^2,\:v'=e^u\\ \int \:wv'=wv-\int \:w'v\\ {/eq}

We get the integral as:

{eq}\int _0^{-3t}-\frac{e^uu^2}{27}du=-\frac{1}{27}\left[u^2e^u-\int \:2ue^udu\right]^{-3t}_0\\ =-\frac{1}{27}\left[u^2e^u-2\left(e^uu-e^u\right)\right]^{-3t}_0\\ => S =-\frac{1}{27}\left(9e^{-3t}t^2-2\left(-3e^{-3t}t-e^{-3t}\right)-2\right)\\ {/eq}


Learn more about this topic:

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Word Problems Involving Time, Distance & Velocity

from CBASE: Practice & Study Guide

Chapter 8 / Lesson 5
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