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Standard IX

Physics

Acceleration

A particle st...

Question

A particle starts at the origin and move along the x axis such that its velocity $v$ varies with time $t$ as $v = 4 t^{3} - 2 t$ , where v is in m/s and t is in second.What is the distance of the particle from the origin when its acceleration is $46 m / s^{2}$ ?

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Solution

$a = \frac{d v}{d t} = 12 t^{2} - 2$
As $a = 46$
$∴ 12 t^{2} - 2 = 46$
$t^{2} = 4$
$t = 2 s e c$
Now, $S = \int_{0}^{2} V d t$
$= \int_{0}^{2} (4 t^{3} - 2 t) d t$
$= [\frac{4 t^{4}}{4} - \frac{2 t^{2}}{2}]_{0}^{2}$
$= 2^{4} - 2^{2} = 16 - 4 = 12 m$

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A particle starts at the origin and move along the x axis such that its velocity v varies with time t as v=4t3−2t, where v is in m/s and t is in second.What is the distance of the particle from the origin when its acceleration is 46m/s2?

a=dvdt=12t2−2As a=46 ∴12t2−2=46 t2=4 t=2secNow, S=∫20Vdt=∫20(4t3−2t)dt=[4t44−2t22]20=24−22=16−4=12m

A particle starts at the origin and move along the x axis such that its velocity $v$ varies with time $t$ as $v = 4 t^{3} - 2 t$ , where v is in m/s and t is in second.What is the distance of the particle from the origin when its acceleration is $46 m / s^{2}$ ?

$a = \frac{d v}{d t} = 12 t^{2} - 2$
As $a = 46$
$∴ 12 t^{2} - 2 = 46$
$t^{2} = 4$
$t = 2 s e c$
Now, $S = \int_{0}^{2} V d t$
$= \int_{0}^{2} (4 t^{3} - 2 t) d t$
$= [\frac{4 t^{4}}{4} - \frac{2 t^{2}}{2}]_{0}^{2}$
$= 2^{4} - 2^{2} = 16 - 4 = 12 m$