A particle moves in a straight line so that s=t, then its acceleration is proportional to

A
${(v e l o c i t y)}^{3}$
B
velocity
C
${(v e l o c i t y)}^{2}$
D
${(v e l o c i t y)}^{\frac{3}{2}}$

Solution:

We have,

$\begin{array}{l} s = \sqrt{t} \Rightarrow \frac{d s}{d t} = \frac{1}{2 \sqrt{t}} and \frac{d^{2} s}{d t^{2}} = - \frac{1}{4 t^{3 / 2}} \\ \Rightarrow \frac{d^{2} s}{d t^{2}} = - \frac{1}{4} {(\frac{2 d s}{d t})}^{3} = - 2 {(\frac{d s}{d t})}^{3} \end{array}$

Hence, Acceleration $\infty$ ${(V e l o c i r y)}^{3}$

A particle moves in a straight line so that $s = \sqrt{t},$ then its acceleration is proportional to

Solution:

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