The distance moved by the particle in time t is given by x=t3−12t2+6t+8

Discussion Forum : Application Of Derivatives

Question -

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

Options:

A . 42

B . -42

C . 48

D . -48

Answer: Option B
:
B
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.

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