A speeder passes a parked police car at a constant speed of 24.8 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.45 m/s2.
How much time passes before the speeder is overtaken by the police car?
Well, this seems like a classic case of "The Tortoise and the Hare: The High-Speed Edition." Let's see how the race unfolds.
First, let's find out how long it takes for the police car to catch up with the speeder. We can use a fancy equation called the "suvat" equation to figure that out. In this case, s stands for the distance traveled, u stands for the initial velocity, v stands for the final velocity, a stands for the acceleration, and t stands for time.
Given that the speeder has a constant velocity of 24.8 m/s and the police car starts from rest (initial velocity = 0) with an acceleration of 2.45 m/s², we're interested in figuring out how much time it takes for the two velocities to be equal.
Using the equation v = u + at, we can rearrange it to solve for time (t):
t = (v - u) / a.
Plugging in the values: v = 24.8 m/s, u = 0 m/s, and a = 2.45 m/s², we get:
t = (24.8 m/s - 0 m/s) / 2.45 m/s².
Crunching the numbers, we find:
t ≈ 10.12 seconds.
So, it will take approximately 10.12 seconds for the police car to catch up with the speeder, assuming nothing extraordinary happens during the chase. I hope everyone involved enjoys the thrilling pursuit with lots of flashing lights and siren symphonies!