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Standard XII

Physics

Differentiation

The distance ...

Question

The distance travelled by a particle is directly proportional to $t^{1 / 2}$ , where 't' is time elapsed. What is the nature of motion?

Increasing acceleration

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Decreasing acceleration

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Increasing retardation

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Decreasing retardation

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Solution

The correct option is D Decreasing retardation
$s = k t^{2}$
$v = \frac{d s}{d t} = \frac{1}{2} k t^{- 1 / 2}$
Acceleration $= \frac{d^{2} s}{d t^{2}} = - \frac{1}{4} k t^{- 3 / 2}$ =negative
With increase in time ,acceleration decreases.
Decreasing retardation.

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The distance travelled by a particle is directly proportional to t1/2, where 't' is time elapsed. What is the nature of motion?

The correct option is D Decreasing retardations=kt2v=dsdt=12kt−1/2Acceleration =d2sdt2=−14kt−3/2=negativeWith increase in time ,acceleration decreases.Decreasing retardation.

The distance travelled by a particle is directly proportional to $t^{1 / 2}$ , where 't' is time elapsed. What is the nature of motion?

The correct option is D Decreasing retardation
$s = k t^{2}$
$v = \frac{d s}{d t} = \frac{1}{2} k t^{- 1 / 2}$
Acceleration $= \frac{d^{2} s}{d t^{2}} = - \frac{1}{4} k t^{- 3 / 2}$ =negative
With increase in time ,acceleration decreases.
Decreasing retardation.