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, the velocity is
(a) 42
(b) $-42$
(c) 48
(d) $-48$
Solution:
(b) $-42$
$x=t^{3}-12 t^{2}+6 t+8$
$\Rightarrow \frac{d x}{d t}=3 t^{2}-24 t+6$
$\Rightarrow \frac{d^{2} x}{d t^{2}}=6 t-24$
$\Rightarrow 6 t-24=0 \quad[\because$ acceleration is zero $]$
$\Rightarrow t=4$
$\mathrm{So}$,
Velocity at $t=4$
$\Rightarrow \frac{d x}{d t}=3(4)^{2}-24 \times 4+6$
$\Rightarrow \frac{d x}{d t}=48-96+6$
$\Rightarrow \frac{d x}{d t}=-42$
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