Ah yes i see now thank you!
However I did ignore +c when I intergrated as I couldn't see how you would determine it since the question doesnt say it starts at rest or anything.
Doing this I get the answer in the mark scheme but I was wondering if you know how you'd know c is 0 or was it just an oversight in the question?
Ah, yes, I failed to use Newton's 2nd law...the ODE should in fact be:
[MATH]3000\d{v}{t}=6000-15v^2[/MATH]
[MATH]200\d{v}{t}=20^2-v^2[/MATH]
I would continue as follows:
[MATH]5\left(\frac{1}{v+20}-\frac{1}{v-20}\right)\,dv=dt[/MATH]
Integrate, using the boundaries as limits and switch the dummy variables:
[MATH]5\int_5^v \frac{1}{y+20}-\frac{1}{y-20}\,dy=\int_0^t\,dz[/MATH]
[MATH]5\ln\left(\frac{3}{5}\left|\frac{v+20}{v-20}\right|\right)=t[/MATH]
And so, to answer the first question, we find:
[MATH]t(15)=5\ln\left(\frac{21}{5}\right)\approx7.175[/MATH]
This agrees with your result, where you have taken \(v(0)=0\). Under that model, the constant of integration would indeed be zero, but you don't want to, in practice, make that assumption.
To answer the second question, I would first get velocity \(v\) as a function of time \(t\)...can you proceed?
