The kinetic energy of a body moving along a s | vedclass

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Question

$\frac{1}{2} m v^{2} \propto t$

$\mathrm{v} \propto \sqrt{\mathrm{t}}$

$\Rightarrow \mathrm{v}=\alpha \sqrt{\mathrm{t}}$

$a=\frac{d v}{d t}=\

Vedclass · Accepted Answer

inversely proportional to velocity.

Vedclass · Answer

$\frac{1}{2} m v^{2} \propto t$

$\mathrm{v} \propto \sqrt{\mathrm{t}}$

$\Rightarrow \mathrm{v}=\alpha \sqrt{\mathrm{t}}$

$a=\frac{d v}{d t}=\frac{\alpha}{2} t^{-1 / 2}=\frac{\alpha}{2 \sqrt{t}}$

$F=m a=\frac{m \alpha}{2 \sqrt{t}}$

$\mathrm{F} \propto \frac{1}{\mathrm{v}}$

The kinetic energy of a body moving along a straight line varies with time as shown in figure. The force acting on body

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