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Standard XII

Physics

Acceleration

The position ...

Question

The position of a body is given as a function of time by the relation, $x = 2 t^{3} - 6 t^{2} + 12 t + 6$ . The acceleration of the body is zero, after a time

1 s

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2 s

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2 \sqrt{2}

s

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3 s

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Solution

The correct option is B $1 s$
$a = \frac{d^{2} x}{d t^{2}} = \frac{d^{2}}{d t^{2}} (2 t^{3} - 6 t^{2} + 12 t + 6) = 0$
$\Rightarrow 12 t - 12 = 0$
$[t = 1 s e c]$

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The position of a body is given as a function of time by the relation, x=2t3−6t2+12t+6. The acceleration of the body is zero, after a time

The correct option is B 1 sa=d2xdt2=d2dt2(2t3−6t2+12t+6)=0⇒12t−12=0 [t=1sec]

The position of a body is given as a function of time by the relation, $x = 2 t^{3} - 6 t^{2} + 12 t + 6$ . The acceleration of the body is zero, after a time

The correct option is B $1 s$
$a = \frac{d^{2} x}{d t^{2}} = \frac{d^{2}}{d t^{2}} (2 t^{3} - 6 t^{2} + 12 t + 6) = 0$
$\Rightarrow 12 t - 12 = 0$
$[t = 1 s e c]$