Speeding Driver & Cop Car Velocity/Time/Acceleration Q

  • Thread starter chrisweikert
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In summary, the conversation is about a makeup exam question involving a speeding driver and a pursuing police car. The variables and equations for solving the problem are also provided. The time it takes for the police car to catch the speeder is 12.3 seconds, the distance the speeder has gone when caught is 565.8 meters, and the speed of the police car when it catches the speeder is 92.3 meters per second. The suggestion is also given to round off the answers after completing the computations.
  • #1
chrisweikert
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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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  • #2
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.
 

1. What is the relationship between a speeding driver's velocity and time?

The relationship between a speeding driver's velocity and time can be described by the equation velocity = distance/time. This means that the velocity of a speeding driver will increase as time passes, as long as the distance traveled remains constant.

2. How is acceleration related to a speeding driver's velocity?

Acceleration is the rate of change of velocity over time. Therefore, a speeding driver's velocity will continue to increase as long as they maintain a constant acceleration. However, if the acceleration changes, the velocity will also change accordingly.

3. Can a speeding driver's velocity be negative?

Yes, a speeding driver's velocity can be negative. This indicates that the driver is moving in the opposite direction of the positive direction, which is usually considered as the direction of motion.

4. How does a cop car's velocity compare to a speeding driver's velocity?

In most cases, a cop car will have a greater velocity than a speeding driver, as they are likely traveling at a higher speed in order to catch the speeding driver. However, this may vary depending on the specific circumstances of the situation.

5. How can the time it takes for a cop car to catch a speeding driver be calculated?

The time it takes for a cop car to catch a speeding driver can be calculated by using the equation time = distance/velocity. The distance can be determined by the initial distance between the two vehicles, and the velocity can be determined by the speed of the cop car. However, this calculation may be affected by other factors such as traffic conditions and the acceleration of both vehicles.

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