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Q.

A particle moves along a straight line $O X$ . At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from O is given by $x = 50 + 12 t - t^{3}$ . How long would the particle travel before coming to rest ?

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a

$66 m$

b

$56 m$

c

$26 m$

d

$16 m$

answer is D.

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Detailed Solution

$x = 50 + 12 t - t^{3} v = \frac{d x}{d t} = 0 + 12 - 3 t^{2} f o r v = 0 \Rightarrow 12 - 3 t^{2} = 0 \Rightarrow t^{2} = 4 \Rightarrow t = 2 s e c \Rightarrow x_{t = 0} = 50 m x_{t = 2 S} = 50 + 12 \times 2 - {(2)}^{3} = 66 d i s \tan c e t r a v e l l e d = x_{t = 2} - x_{t = 0} = 66 - 50 = 16 m$

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