Question

A speeder passes a parked police car at a constant speed of 36.7m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.83m/s. How much time passes before the speeder is overtaken by the police car? how far does the speeder travel before being overtaken by the police car in meters

Answer 1

So the speeder has position

x = 36.7t

the police car has position

x = (1/2)gt^2

(where g is the acceleration of the police car). when the police car is overtaken, their x positions are the same so we can equate them

36.7t = x = (1/2)gt^2

now solve this for t by dividing t from each side

36.7 = (1/2)gt
(2 x 36.7)/g = t
t = 25.9 seconds

to find the distance they will be, plug this t back into either equation. ill use the first

x = 36.7t = 36.7 x 25.9 = 950m