The distance moved by the particle in time t is given b

The distance moved by the particle in time t is given by $x = t^{3} - 12 t^{2} + 6 t + 8$ . At the instant when its acceleration is zero, then the velocity is

-42

-48

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Right Answer is: B

SOLUTION

We have,
$x = t^{3} - 12 t^{2} + 6 t + 8$
$\Rightarrow \frac{d x}{d t} = 3 t^{2} - 24 t + 6 a n d \frac{d^{2} x}{d t^{2}} = 6 t - 24$
Now, Acceleration =0
$\Rightarrow \frac{d^{2} x}{d t^{2}} = 0 \Rightarrow 6 t - 24 = 0 \Rightarrow t = 4$
Att=4, we have
Velocity = ${(\frac{d x}{d t})}_{r - 4} = 3 \times 4^{2} - 24 \times 4 + 6 = - 42$ .
Hence (b) is the correct answer.

The distance moved by the particle in time t is given b

Right Answer is: B

SOLUTION

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