Question

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 45.0 ft to stop. If the same motorcycle is traveling 73.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration.

Answer 1

First case,

v1= 44 mi/h= 64.53 ft/s

s1= 45 ft

Using Kinematics 3rd equation of motion

a= v1^2/(2*s1)= 64.53^2/(2*45)

a= 46.27 ft/s^2

Now

v2 = 73 mi/h

v2= 107.067 m/s

s2= v2^2/(2* 46.27)

s2= 123.87 ft