An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 135 km/h. precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.60 m/s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)? 12.

 Timing of catch up = 9.45 seconds after acceleration of police car.

Explanation:

Speed of police car = 95 km/h = 95 x 5 / 18 = 26.39 m/s

Speed of speeder = 135 km/h = 135 x 5 / 18 = 37.5 m/s

Let the time of catch up be t seconds.

Distance travelled by police car must be equal to distance travelled by speeder.

Distance travelled by speeder = Speed x Time = 37.5 x ( t + 1)

Distance travelled by police car is found out by , s is the displacement, u is the initial velocity, a is the acceleration and t is the time after 1 second.

In this case the velocity of police car in vertical direction = 26.39 m/s, acceleration = 2.60  and time is t.

Distance travelled by police car

Equating both distances,

Negative time is not possible,

       So time is t = 9.45 seconds

Timing of catch up = 9.45 seconds after acceleration of police car.

do you have a formula possibly