Here is a problem I been stuck on for a while, it seems like there is a simple formula that i am missing. I've tried to use physics applications to this but I'm not sure I'm using the right formula. In need of the steps in order to do this problem, please!
A motorboat starts from rest. Its motor provides a constant acceleration of $5 \frac {\mathbf{ft}}{\mathbf{s}^2}$ while water resistance causes a deceleration of $\frac {v}{10} \frac{\mathbf{ft}}{\mathbf{s}^2}$ (v is velocity). Find $v$ when $t=10 \mathbf{s}$.
So we have $v'(t) = 5 - \frac {v(t)}{10}$ and $v'(0) = 0$. This is an inhomogenous, first degree differential equation with constant coefficients. We could solve it directly, but it will get a bit easier to solve if we rewrite it to $$ [v(t)-50]' = -\frac{v(t) - 50}{10} $$ which means that $w(t) = v(t)-50$ fulfills $w'(t) = -\frac{w(t)}{10}$. Together with $w(0) = -50$, we therefore have $w(t) = -50e^{-t/10}$. Transforming back into $v$, we get $$ v(t) = w(t) + 50 = -50e^{-t/10} + 50 $$ Lastly, we were asked to calculate $v(10)$, which is equal to $$ v(10) = -50e^{-10/10} + 50 = -50e^{-1} + 50 \approx 31.6 $$