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Question

$s=k t^{1 / 2}$

$\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{dt}^{2}}=-\frac{1}{4} \mathrm{kt}^{-3 / 2}$

As $t$ increases, the retardation decrease

Vedclass · Accepted Answer

Decreasing retardation

Vedclass · Answer

$s=k t^{1 / 2}$

$\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{dt}^{2}}=-\frac{1}{4} \mathrm{kt}^{-3 / 2}$

As $t$ increases, the retardation decreases

The distance travelled by a particle is directly proportional to $t^{1/2}$, where $t =$ time elapsed. What is the nature of motion ?

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