Initial Velocity

A motorcycle is accelerating at a constant rate of 5 ft/sec^2. It travels 100 ft while its speed changes to 60 ft/s. What is the initial velocity in ft/s?

Expand Hint
$$$v^2=v_0^2+2a_0(s-s_0)$$$
where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ .
Hint 2
Solve for $$v_0$$
For constant acceleration, the equation for velocity as a function of position:
$$$v^2=v_0^2+2a_0(s-s_0)$$$
where $$v$$ is the velocity along the direction of travel, $$v_0$$ is the velocity at time $$t_0$$ , $$a_0$$ is constant acceleration, $$s$$ is the displacement at time $$t$$ along the line of travel, and $$s_0$$ is the displacement at time $$t_0$$ .

Solving for initial velocity:
$$$(60ft/s)^2=v_0^2+2(5\frac{ft}{s^2})(100ft)$$$
$$$3,600ft^2/s^2=v_0^2+1,000ft^2/s^2$$$
$$$v_0=\sqrt{(3,600ft^2/s^2-1,000ft^2/s^2)}=\sqrt{2,600ft^2/s^2}=51\:ft/s$$$
51 ft/s