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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The position of a particle moving along the $x$-axis ($x$ is in $m$) is given by $x = t^3 - 3t^2 + 2t$. Find the average speed of the particle in the interval $t=2\;s \rightarrow t = 4\;s$

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Answer: 12 m/s
Given $x = t^3 - 3t^2 + 2t$.
Average Velocity $v_{\text{avg}} = \large\frac{\text{displacement}}{\Delta \text{time}}$
$\Delta$ time $= 4s - 2s = 2s$
Position at $t = 2s \rightarrow x(2) = 2^3 - 3\times 2^2 + 2\times2 = 0$
Position at $t = 4s \rightarrow x(4) = 4^3 - 3\times 4^2 + 2\times4 = 24\;m$
$\Rightarrow$ Displacement $ = 24 - 0 = 24\;m$
$\Rightarrow$ Average velocity $ = \large\frac{24}{2} $$ = 12\;m/s$
answered Aug 12, 2014 by balaji.thirumalai
edited Aug 12, 2014 by balaji.thirumalai
 

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