Question

A top sprinter reached his maximum instantaneous velocity of 12.28 meters per second 7.17 seconds after leaving the blocks. Calculate his acceleration. All key steps must be shown.

I am very confused will you please help me!!

Answer 1

Here we will assume two things.

His acceleration is constant until he reaches his maximum velocity and then he stops accelerating.

When he leaves the blocks, his velocity is "zero", meaning that the initial velocity is zero.

Then:

For the acceleration of the sprinter we will have a constant, A:

a(t) = A.

for the velocity we need to integrate over time, we will get

v(t) = A*t + v0

Where v0 is a constant of integration, and this is the initial velocity, as we already said, this will be zero, so the velocity equation is:

v(t) = A*t

We know that at t = 7.17 seconds, his velocity is 12.28m/s

then:

v(7.17s) = 12.28m/s = A*7.17s

12.28m/s/7.17s = 1.71 m/s^2 = A.

The acceleration is: 1.71 m/s^2 for 0 < t < 7.17s

0 m/s^2 for 7.17s < t.

This is because that was his maximum velocity, then after that time he can not accelerate more.