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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 $t=10 \mathbf{s}$.

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    Whould there be a $v$ in the expression for the water resistance? – 2017-02-02
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    yes there would be a v for the water resistance – 2017-02-02
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    The formula you gave for water resistance doesn't have a v value in it. Did you copy the problem correctly? – 2017-02-02
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    Oh, i see what you're talking about, the v in this problem was talking about the deceleration caused by the water resistance, which was equal to 1/10v ft/sec^2 – 2017-02-02
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    technically you can say the deceleration is also equivalent to v/10 ft/sec^2 – 2017-02-02

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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 $$ v(10) = -50e^{-10/10} + 50 = -50e^{-1} + 50 \approx 31.6 $$