## 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