The Physics Behind Pit Stops
This page relates the equations of kinematics for constant acceleration to a pit stop.
Average Velocity = xf - xi / tf - ti
Average Acceleration = vf - vi / tf - ti
Equations of Kinematics:
v = vo + at
x = 1/2(vo + v)t
x = vot + 1/2at2
v2 = vo2 + 2ax
Problem:
Jeff Gordon traveling at a constant speed of 175 mph passes Dale Earnhardt Jr.'s pit stop, at which point Earnhardt Jr. accelerates from rest. How much acceleration does Earnhardt Jr. need to meet Gordon's car by the time he reaches the end of his pit stop? Pit stop = 20m
Solution:
We know Gordon's v = 175mi/hr, but we need to convert it to m/s:
175 mi/hr * 1.609km/1mi * 1000m/1km * 1hr/3600s = 78.3 m/s
Gordon is traveling the distance x = 20m, at a constant acceleration. He reaches x = 20m at time t. We need to use the equation:
x = vot + 1/2at2 to find t.
Because the a = 0 (constant), 1/2at2 = 0, we can use the equation:
x = vot
20m = 78.2m/s(t)
t = .256s
Earndardt Jr. wants to reach the same distance, x=20, as Gordon at the same time, t = .256s...we need to find his acceleration.
We need to use the equation:
x = vot + 1/2at2
Because he is starting from rest, his initial velocity, vo=0. Therefore,
vot = 0. We can use the equation:
x = 1/2at2
20m = 1/2(a)(.256s)2
a= 610.35 m/s2
Earnhardt must accelerate at 610.35 m/s2 !
To learn more about the equations of kinematics for constant acceleration go to: