Question

A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44 m/s².
a. How much time passes before the speeder is overtaken by the police car?
b. How far does the speeder get before being overtaken by the police car?

Answer 1

Answer

given,

speed of the car = 30 m/s

initial speed of the police car = 0 m/s

acceleration of the car = 2.44 m/s²

time taken by the car to overtake the speeder

this will occur when the distance traveled by the police and the speeder is same.

Distance by the speeder = s x t = 30 x t-----(1)

distance traveled by the police

$s = ut + \dfrac{1}{2}at^2$

$s = \dfrac{1}{2}at^2$-----------(2)

equating both the equation

$30 t = \dfrac{1}{2} \times 2.44 t^2$

30 = 1.22 t

t = 24.6 s

b) distance covered by the speeder

   from equation (1)

 D = 30 x t

 D = 30 x 24.6

 D = 737.7 m

Distance travel by the speeder = 737.7 m.