Question

A car stops from 60 mph in a distance of 155 feet. (The tires do not spin or slide. The tire diameter is 24 inches.)

a) For the tires, what is the angular acceleration in revolutions per second squared?

b) Through how many revolutions did it turn?

Answer 1

given
vi = 60 mph

= 60*1609/(60*60) m/s

= 26.82 m/s

vf = 0

d = 155 feet

= 155*0.3048

= 47.2 m

radius of tghe tire, r = 24/2

= 12 inch

= 1 feet

= 0.3048 m

a) let a is the acceleration of the car, a = (vf^2 -vi^2)/(2*d)

= (0^2 - 26.82^2)/(2*47.2)

= -7.62 m/s^2

angular acceleration, alfa = a*r

= -7.62*0.3048

= -2.32 rad/s^2

= -2.32/(2*pi) rev/s^2

= -0.369 rev/s^2 <<<<<<<-----------------Answer

b) no of revolutions, N = d/(2*pi*r)

= 155/(2*pi*1)

= 25 <<<<<<<-----------------Answer