Given a(t) = -Cv^2(t) where C is a constant, express v(t) and x(t) as explicit functions of time t. Assume v(0) = v0(adsbygoogle = window.adsbygoogle || []).push({});

So I tried integrating both sides of the equation to get:

v(t)+C' = -C integral v^2(t) dt

but then I how am I supposed to integrate v^2(t) dt...

Also, I thought I could do it another way:

velocity

a(t) = -Cv^2(t)

v(t)/t = -Cv^2(t)

-1/(Ct) = v(t)

distance

-1/(Ct) = v(t)

-1/(Ct) = x(t)/t

-1/C = x(t)

Well the distance can't be constant so I guess that doesn't work either. I don't even know why this 2nd method doesn't work... It looks fine to me. Any help would be appreciated.

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# Derive Velocity Equation

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