Answer
A constant acceleration of $5.87~ft/s^2$ is required.
Work Step by Step
We can express $30~mi/h$ in units of $ft/s$:
$(30~mi/h)\times \frac{5280~ft}{1~mi}\times \frac{1~h}{3600~s} = 44~ft/s$
We can express $50~mi/h$ in units of $ft/s$:
$(50~mi/h)\times \frac{5280~ft}{1~mi}\times \frac{1~h}{3600~s} = 73.33~ft/s$
$a(t) = \frac{dv}{dt} = a$
$v(t) = v_0+a~t$
When $t = 0$, then $v = 44~ft/s$
Thus $v_0 = 44~ft/s$
$v(t) = 44+a~t$
We can find $a$:
$v(5) = 44+a~(5) = 73.33$
$5a = 29.33$
$a = 5.87~ft/s^2$
A constant acceleration of $5.87~ft/s^2$ is required.
