A sprinter crosses the finish line

Question

A sprinter crosses the finish line at 21.3 m/s and slows down for 1.2 seconds over an 18.2 m period. Assuming uniform acceleration, what was the sprinters final velocity at the end?

Answer 1

So if we have constant acceleration, then:

x(t) = x_i + v_i * t + 1/2 * a * t^2

where x_i is the initial position and v_i is the initial velocity. We know that x(t=1.2) = 18.2 m and v_i = 21.3 m/s. If we can use these numbers to solve for the acceleration. Let t_f = 1.2s and x_f = x(1.2) = 18.2m. Then, we find that

x_f = x_i + v_i * t_f + 1/2 * a * t_f

(x_f - x_i) = t_f * (v_i + 1/2 * a * t_f)

v_i + 1/2 * a * t_f = (x_f - x_i) / t_f

1/2 * a * t_f = (x_f - x_i) / t_f - v_i

a = (2 / t_f) * ((x_f - x_i) / t_f - v_i)

a = (2 / 1.2) * (18.2 / 1.2 - 21.3)

a = -10.2 m/s^2

Now that we have the acceleration, we can compute the final velocity:

v(t) = v_0 + a * t

v(t=1.2) = 21.3 + (-10.2) * 1.2

= 9.03m/s