Question

An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the police officer steps on the accelerator. If the police car accelerates uniformly at 3.00 m/s2 and overtakes the speeder after accelerating for 5.00 s , what was the speeder's speed?

Answer 1

Initial constant speed of the police car = V₁ = 85 km/hr = 85 x (1000/3600) m/s = 23.61 m/s

Time period for which the police car travels at same speed = T₁ = 2 sec

Distance traveled by the police car at same speed = D₁

D₁ = V₁T₁

D₁ = (23.61)(2)

D₁ = 47.22 m

Acceleration of the police car = a = 3 m/s²

Time period the police car accelerates for = T₂ = 5 sec

Distance traveled by the police car while accelerating = D₂

D₂ = V₁T₂ + aT₂²/2

D₂ = (23.61)(5) + (3)(5)²/2

D₂ = 155.55 m

Speed of the speeder car = V

Distance traveled by the speeder car = D

D = V(T₁ + T₂)

The total distance traveled by the police car and the speeder car are equal.

D = D₁ + D₂

V(T₁ + T₂) = D₁ + D₂

V(2 + 5) = 47.22 + 155.55

V = 28.97 m/s

Speed of the speeder car = 28.97 m/s