Question

A police officer in a patrol car parked in a 70 Km/h speed zone observes a passing automobile might be intoxicated,the officer starts his car,accelerates uniformly to 90 Km/h in 8s ,and maintaining a constant velocity of 90 Km/h,overtakes the motorist 42s after the automobile passed him. Knowing that 18s elapsed before the officer began pursuing the motorist,determine
A)The distance the officer travelled before overtaking the motorist
B)The motorist speed

Answer 1

90km/h = 90,000m/3600s = 25 m/s.

a = (V-Vo)/t = (25-0)/8 = 3.125 m/s^2.

d1 = 0.5a*t^2 = 0.5*3.125*8^2 = 100 m. =
distance traveled while accelerating.

t = 42-18 = 24 s. Driving time.

A. d = 100 + 25m/s * (24-8)s. = 500 m.

B. d = Vm*t = 500 m.
V*42 = 500.
V = 11.9 m/s. = 43 km/h.