Precisely 1.50 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20 m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)

11.42 s

Explanation:

Assuming the speed of the police car = 99 Km/h

       v = 99 km/h x 0.278 = 27.52 m/s

Speed of the Passer = 133 Km/h

       u = 133 x 0.278 = 37 m/s

acceleration of the police car = 2.20 m/s²

time taken to overtake passer = ?

distance travel by the passer in t second.

x = u t

x = 37 t

Distance traveled by the police

both the distance are same so,

37 t = 27.52 t +1.1 (t-1.5)²

8.62 t = t² + 2.25 - 3t

t² -11.62 t + 2.25 = 0

on solving the above equation we get

t = 11.42 s , 0.92 s

neglecting the second value because it is less than 1.5 s.

Time taken by the driver to overtake the passer is equal to 11.42 s

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