A particle moves along a straight line $OX$. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^{3}$ How long would the particle travel before coming to rest

Question

Dr. Abubacker · Accepted Answer

(c) Distance travelled by the particle is
$x = 40 + 12t - t^{3}$
We know that, speed is rate of change of distance i.e.
$v = \frac{dx}{dt}$
$	herefore v = \frac{d}{dt}\left( 40 + 12t - t^{3} ight) = 0 + 12 - 3t^{2}$
But final velocity $v = 0$
$	herefore 12 - 3t^{2} = 0$
$\Rightarrow t^{2} = \frac{12}{3} = 4 \Rightarrow t = 2s$
Hence, distance travelled by the particle before coming to rest is given by
$x = 40 + 12(2) - (2)^{3} = 40 + 24 - 8 = 64 - 8 = 56m$

Q. A particle moves along a straight line $OX$. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^{3}$ How long would the particle travel before coming to rest

Explanation

Reviewed by Dr. Abubacker

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