Answer to Question #87000 in Mechanics | Relativity for Ester

Question

While driving along a road an overloaded truck traveling at 79 km h−1 passes a stationary police car at the side of the road.
The police car begins to accelerate at a constant rate until it catches up and passes the truck 2.8 km down the road.
What is the acceleration required for the the police car to catch the truck in this distance? (in m s−2 to 2 s.f)

Answer 1

Velocity of truck(v)=79 km/h

Initial velocity of police car (u)=0

Distance travelled by police car(s)=2.8km

Assuming truck is moving at a constant velocity. So the time required to travel 2.8km =t

We know, s=vt,So,t=s/v ; t=2.8km/79km/h =0.0354h

t=0.0354h

Again police car is moving at a constant accelearation=f

We know, s=ut +(1/2)ft²

s=(1/2)×ft2,since u=0

f=2s/t² = (2× 2.8)/(0.0354)² km/h² =4468.7 km/h²

f=4468.7× 1000m/(3600) ² m/s²

f=0.345 m/s²