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**bigmansouf** Q: if the velocity v is given by the formula$\displaystyle v = \frac{u}{(1+ks)}$ where u is the initial velocity, s is the distance and k is a constant prove that the accleration varies as $\displaystyle v^{3}$

my attempt:

$\displaystyle v \frac{u}{(1+ks)}=\frac{a}{b}$

$\displaystyle \frac{\mathrm{d}v}{\mathrm{d}s}=\frac{b\frac{\math rm{d}a}{\mathrm{d} s}-a\frac{\mathrm{d} b}{\mathrm{d} s}}{b^2}$

$\displaystyle a=u; \frac{\mathrm{d} a}{\mathrm{d} s} =0 $

$\displaystyle b = 1+ks; \frac{\mathrm{d} b}{\mathrm{d} s}= k $

$\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{((1+ks)(0)-u(k))}{(1+ks)^2}b$

$\displaystyle \frac{\mathrm{d} v}{\mathrm{d} s}= \frac{-uk}{(1+ks)^2} $

we know that a = v . dv/ds

thus; $\displaystyle a = \frac{u}{(1+ks)} \times \frac{-uk}{(1+ks)^2} $

$\displaystyle a = \frac{-u^{2}k}{(1+ks)^2}$

note $\displaystyle s = \frac{u-v}{kv} $ sub s into a therefore;

$\displaystyle a = -\frac{kv^3}{u}$

**thus as v tend to infinity, $\displaystyle v^{3} $ tends to infinity thus acceleration varies as $\displaystyle v^{3} $**

I want to know if I am right?

Also the bold part is the correct reasoning?