Question

When starting a foot race, a 61.5 kg sprinter exerts an average force of 665 N backward on the ground for 0.75s.

Randomized Variables

m = 61.5 kg

f = 665 N

t = 0.75 s

What is his final speed in m/s?

How far does he travel in meters?

Answer 1

Answer:

8.10975 m/s

3.04115625 m

Explanation:

t = Time taken = 0.75 s

u = Initial velocity = 0 (starting from rest)

v = Final velocity

s = Displacement

a = Acceleration

F = Force = 665 N

m = Mass of person = 61.5 kg

Force is given by

$F=ma\\\Rightarrow a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{665}{61.5}\\\Rightarrow a=10.813\ m/s^2$

From equation of motion

$v=u+at\\\Rightarrow v=0+10.813\times 0.75\\\Rightarrow v=8.10975\ m/s$

The person's final speed is 8.10975 m/s

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 10.813\times 0.75^2\\\Rightarrow s=3.04115625\ m$

The distance the person traveled is 3.04115625 m