A police car at rest chases a passing car which has a velocity of 80 mph. The time required for the police car to catch the passing car needs to be determined.

The velocity function of both cars are shown on the left. Transfer the unit of velocity to miles/minute to get better graphics. Then the velocity of the passing car is 1.333 miles/minute , and the police car's velocity function is
      
where v is in miles/min, and t is in minute. When the police car reaches the passing car, the distance both car travel will be the same. That is, the area under the velocity curves for each car will be the same. So the problem becomes, when the blue region and the purple region has the same area, the police car catches the passing car.

Velocity is simply v = ds/dt, so the distance, s, can be determined as,

Assume t1 is the time the police car's velocity reaches the velocity of the passing car (v = 1.333 mile/min), t₂ is the time the police car's velocity reaches 120 mph (2 mile/min), and t₃ is the time the police car reaches the passing car.

Simplifying the above equation gives,

First, t₂ can be determined since the maximum velocity of the police car is known.

      v = 2.0 = 0.4t₂ + t₂ ^0.5
      t₂ = 1.721 min

Next, integrate the equation and solve for t₃ gives

      1.333t₂ - 0.2t₂² - 2/3(t₂)^1.5 - 0.6667(t₃ - t₂) = 0
      1.333(1.721) - 0.2(1.721)² - 2/3(1.721)^1.5
      - 0.6667(t₃ - 1.721) = 0
      t₃ = 2.016 min

The result shows the police car needs 2.016 minutes to catch the passing car.