Write two equations for distance travelled vs time after the speeder passes. One equation for each car.
Set the distances equal and solve for the time t.
V1 = 120 km/h = 33.33 m/s is the speeder's speed
V2 = 26.39 m/s + 1.90 (t-1)
is the policeman's speed,for t>1 s. For 0<t<1, it is 26.39 m/s
X1 (the speeder) = 33.33*t
X2 (the cop) = 26.39*t + 0.95(t-1)^2
Use X1 = X2 to solve for t
An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 120 km/h.
Precisely 1.00 sec. after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
4 answers
6.1 seconds
6.1
I did not understand
2 answers