Question

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Answer 1

We know that acceleration = dv/dt
Therefore,
 $\frac{dv}{dt} =k x^{2}$
now, multiply the ohs and res with dx,
 $dx \frac{dv}{dt} =k x^{2} dx$
we know that dx/dt= velocity
therefore,
 $vdv=k x^{2} dx$
Now integrate:
 $\int\limits {v} \, dv=k \int\limits { x^{2} } \, dx$
 $\frac{ v^{2} }{2}=k \frac{ x^{3} }{3}+c$

Hence, v is directly proportional to $x^{3}$