Speeding Driver & Cop Car Velocity/Time/Acceleration Q

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The discussion revolves around a physics problem involving a speeding driver and a police car. The driver maintains a speed of 46 m/s while the police car accelerates at 7.5 m/s² to catch up. The calculations show that it takes approximately 12.3 seconds for the police car to catch the speeder, during which the speeder travels about 565.8 meters. When the police car catches the speeder, its speed reaches approximately 92.3 m/s. The importance of rounding only after completing calculations is emphasized for accuracy.
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So I'm new here, and I apologize if I've posted in the wrong format/place/etc., but this is a makeup exam question, and I (think) I've finished it, but I want a second opinion on the correctness if you guys wouldn't mind. :smile:

Homework Statement


A speeding driver passes a hidden police car moving at a speed of 46m/s; assume the speeder keeps going this speed. The police car immediately starts accelerating after the speeder at 7.5m/s2.
a. How long does the police car take to catch the speeder?
b. How far has the speeder gone when police catches him?
c. How fast is the police car moving when it catches the speeder?

Variables for speeder denoted with 's', for police denoted with 'p'.
Vs = 46m/s
as = 0m/s2
t = ?
Δx = ?
ap = 7.5m/s2
Vos = 0m/s
Vfp = ?

2. Homework Equations & 3. The Attempt at a Solution

a. Δx = 46m/st for the speeder & Δx = 1/2(7.5m/s2)t2 for the police car.
So, Δx = 3.75m/s2t2 - 46m/st
Plug into quadratic formula and come up with 12.3s.

b. Δx = 46m/s(12.3s) = 565.8m

c. Vfp = 7.5m/s2(12.3s) = 92.3m/s

Thanks in advance for any comments or helpful information!
 
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I get close to your answers because one should round off after the computation is completed. Don't round off your time, then use the rounded time to compute other results.
 
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