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Acceleration

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Question

The distance moved by the particle in time $^{'} t^{'}$ is given by $S = t^{3} - 12 t^{2} + 6 t + 8$ . At the instant, when its acceleration is zero, the velocity is

42

units/sec

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- 42

units/sec

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48

units/sec

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- 48

units/sec

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Solution

The correct option is B $- 42$ units/sec
$S = t^{3} - 12 t^{2} + 6 t + 8$
$v = \frac{d S}{d t} = 3 t^{2} - 24 t + 6$
$a = \frac{d v}{d t} = 6 t - 24$
$\Rightarrow 6 t - 24 = 0$
$\Rightarrow t = 4$

So, $v (t = 4) = 3 (4)^{2} - 24 (4) + 6 = 3 (16) - 96 + 6$
$\Rightarrow v = - 42$ units/sec

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Acceleration

Standard VII Physics

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The distance moved by the particle in time ′t′ is given by S=t3−12t2+6t+8. At the instant, when its acceleration is zero, the velocity is

The correct option is B −42 units/secS=t3−12t2+6t+8 v=dSdt=3t2−24t+6 a=dvdt=6t−24 ⇒6t−24=0 ⇒t=4 So, v(t=4)=3(4)2−24(4)+6=3(16)−96+6 ⇒v=−42 units/sec

The distance moved by the particle in time $^{'} t^{'}$ is given by $S = t^{3} - 12 t^{2} + 6 t + 8$ . At the instant, when its acceleration is zero, the velocity is

The correct option is B $- 42$ units/sec
$S = t^{3} - 12 t^{2} + 6 t + 8$
$v = \frac{d S}{d t} = 3 t^{2} - 24 t + 6$
$a = \frac{d v}{d t} = 6 t - 24$
$\Rightarrow 6 t - 24 = 0$
$\Rightarrow t = 4$

So, $v (t = 4) = 3 (4)^{2} - 24 (4) + 6 = 3 (16) - 96 + 6$
$\Rightarrow v = - 42$ units/sec