If a particle moving along a line follows the s=1+t, then the accelertion is proportional to

A
square of the velocity
B
cube of the displacement
C
cube of the velocity
D
square of the displacement

Solution:

We have,

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

If a particle moving along a line follows the $s = \sqrt{1 + t}$ , then the accelertion is proportional to

Solution:

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