Question

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.75 m/s 2 . How much time passes before the speeder is overtaken by the police car? Answer in units of s.

Answer 1

It takes approximately 16.95 seconds for the police car, accelerating at 2.75 m/s² from rest, to overtake a speeder traveling at a constant speed of 23.3 m/s.

Step-by-step explanation:

The question asks us to determine how long it will take for a police car starting from rest to overtake a speeder traveling at a constant speed. To solve this, we can use the equations of motion for uniformly accelerated linear motion.

For the speeder (constant velocity):
x = vt

For the police car (uniform acceleration from rest):
x = 0.5at²

We set the two equations equal to each other since they will have traveled the same distance (x) when the police car overtakes the speeder:
vt = 0.5at²

Given that the speeder's velocity (v) is 23.3 m/s and the police car's acceleration (a) is 2.75 m/s², we can solve for time (t):

23.3t = 0.5(2.75)t²
t = 0 (ignored because it's the start time)
t = 2(23.3) / 2.75 ~ 16.95 seconds

Therefore, it takes approximately 16.95 seconds for the police car to overtake the speede