A policeman traveling 16.7m/s (60km/h) spots a speeder ahead, so he accelerates his vehicle at a steady rate of 2.22m/s^2 for 4s, at which time he catches up with the speeder

a) How fast is the police car traveling after 4.00s?
b) How far does the police car travel during the 4.00s?

Question

A policeman traveling 16.7m/s (60km/h) spots a speeder ahead, so he accelerates his vehicle at a steady rate of 2.22m/s^2 for 4s, at which time he catches up with the speeder

a) How fast is the police car traveling after 4.00s?
b) How far does the police car travel during the 4.00s?

anonymous · Answer

2.22m/s^2x4=8.88m/s
8.88m/s+16.7m/s=25.58m/s after 4.00s

anonymous · Answer

s=ut+(1/2)at^2
distance traveled =(16.7*4 ) +(1/2)(2.22)(4^2)
                          =66.8+17.76
                          =84.56m