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Average Rate of Change

The distance ...

Question

The distance moved by a particle in time $t$ seconds is given by $s = t^{3} - 6 t^{2} - 15 t + 12$ . The velocity of the particle when acceleration becomes zero is

15

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- 27

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6 / 5

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none of these

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Solution

The correct option is D $- 27$
$s = t^{3} - 6 t^{2} - 15 t + 12$
Velocity $v = \frac{d s}{d t} = 3 t^{2} - 12 t - 15$
Acceleration $\frac{d v}{d t} = 6 t - 12$
$\Rightarrow 6 t - 12 = 0$
$\Rightarrow t = 2 s e c$
So, $v = - 27$ unit/sec

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