Please solve and answer. Need done ASAP.

A speeder passes a parked police car at 35.0 m/s. The police car starts from rest with a uniform acceleration of 2.39 m/s2.
(a) How much time passes before the speeder is overtaken by the police car? __ s
(b) How far does the speeder get before being overtaken by the police car?

The cover the same distance, in the same time.

You have two equations, one for the police car (starting from rest)
the other the speeder.
set each distance equal, solve for time.
then solve for distance.

To find the time it takes for the police car to overtake the speeder, we can use the formula for calculating time given initial velocity (u), acceleration (a), and displacement (s):

t = (v - u) / a,

where:
t = time,
v = final velocity,
u = initial velocity, and
a = acceleration.

For part (a), the initial velocity of the police car (u) is 0 m/s, the acceleration (a) is 2.39 m/s^2, and the final velocity of the speeder (v) is 35.0 m/s. Plugging these values into the formula, we get:

t = (35.0 - 0) / 2.39.

Calculating this, we find that the time it takes for the police car to overtake the speeder is approximately 14.64 seconds.

For part (b), we need to find the distance the speeder travels in this time. We can use the formula:

s = ut + (1/2)at^2,

where:
s = displacement.

Since the speeder is moving at a constant speed of 35.0 m/s, its initial velocity (u) is also 35.0 m/s. The time (t) is 14.64 seconds, and the acceleration (a) of the police car does not affect the speeder's displacement. Plugging in these values, the formula becomes:

s = (35.0)(14.64).

Calculating this, we find that the speeder travels approximately 511.4 meters before being overtaken by the police car.