Question

A car moving initially at a speed of 52.0 mi/h and weighing 3,000 lb is brought to a stop in a distance of 205 ft. Find (a) the breaking force (b) the time required to stop. Assuming the same braking force find (a) the distance and (d) the time required to stop if the car were going 26.0mi/h initially. (Hint: Convert to SI units)

Answer 1

vi = 52 mi/h

= 52*1609/(60*60) m/s

= 23.2 m/s

vf = 0

d = 205 ft

= 205*0.3048

= 62.5 m

m = 3000 lb

= 3000*0.0.4536

= 1361 kg

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

= (0^2 - 23.2^2)/(2*62.5)

= -4.3 m/s^2

a) Breaking force, F = m*|a|

= 1361*4.3

= 5852 N

b) t = (vf - vi)/a

= (0 - 23.2)/(-4.3)

= 5.40 s

c) vi = 26*1609/(60*60) m/s

= 11.6 m/s

d = (vf^2 - vi^2)/(2*a)

= (0^2 - 11.6^2)/(2*(-4.3))

= 15.6 m

d) t = (vf - vi)/a

= (0 - 11.6)/(-4.3)

= 2.70 s