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Standard XII

Mathematics

Average Rate of Change

The motion of...

Question

The motion of a particle along a straight line is described by equation
$x = 8 + 12 t t^{3}$
where x is in metre and t in second. The retardation of the particle when its velocity becomes zero is

6

^{- 2}

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12

^{- 2}

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24

^{- 2}

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Zero

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Solution

The correct option is B $12$ ms $^{- 2}$
$x = 8 + 12 t - t^{3}$
$v = \frac{d x}{d t} = 12 - 3 t^{2}$
$a = \frac{d v}{d t} = - 6 t$
putting $v = 0$
$3 t^{2} = 12$
$t = 2 s e c$
$a = - 6 \times 2 = - 12 m s^{- 2}$
Retardation $= 12 m s^{- 2}$

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The motion of a particle along a straight line is described by equationx=8+12tt3where x is in metre and t in second. The retardation of the particle when its velocity becomes zero is

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The motion of a particle along a straight line is described by equation
$x = 8 + 12 t t^{3}$
where x is in metre and t in second. The retardation of the particle when its velocity becomes zero is

The correct option is B $12$ ms $^{- 2}$
$x = 8 + 12 t - t^{3}$
$v = \frac{d x}{d t} = 12 - 3 t^{2}$
$a = \frac{d v}{d t} = - 6 t$
putting $v = 0$
$3 t^{2} = 12$
$t = 2 s e c$
$a = - 6 \times 2 = - 12 m s^{- 2}$
Retardation $= 12 m s^{- 2}$