What Is the Correct Formula for Stopping Distance in Physics Homework?

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Homework Help Overview

The discussion revolves around the correct formula for stopping distance in a physics context, specifically involving a car decelerating at a constant rate. The original poster presents multiple statements regarding position, velocity, and stopping distance, seeking clarification on their validity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze various statements about motion equations and stopping distance, questioning the correctness of each. Some participants explore the implications of average versus instantaneous velocity, while others express confusion over the accuracy of the stopping distance equation presented.

Discussion Status

Participants have offered insights regarding the interpretation of the stopping distance equation and its correctness. There is an ongoing examination of the statements provided, with some participants pointing out discrepancies and seeking confirmation on the accuracy of the original problem's wording.

Contextual Notes

There is mention of a practice test created by the professor, which may influence the interpretation of the stopping distance equation. The original poster's transcription of the problem has been questioned, indicating potential issues with the information provided.

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Homework Statement



The driver of a car at 31 m/s sees (t = 0, x = 0) an obstacle down the road and brakes. The car slows down with a constant deceleration which has a magnitude of 6 m/s^2. Select the correct statement below (let d = stopping distance, all units are SI)

a. The position x(t) = 31t + 3t^2
b. The velocity v(t) = 31 - 3t
c. Stopping distance = 961 - 12d
d. The velocity v(t) = d/t
e. None of the above

Homework Equations



Vf = Vi + at
Vf^2 = Vi^2 + 2ad
Xf = Xi + Vi*t + .5at^2

The Attempt at a Solution



a. The position x(t) = 31t + 3t^2

is not correct, because Xf = Xi + Vi*t + .5at^2, so x(t) = 31t - 3t^2

b. The velocity v(t) = 31 - 3t

is not correct, because Vf = Vi + at, so v(t) = 31 - 6t

c. Stopping distance = 961 - 12d

is correct, because Vf^2 = Vi^2 + 2ad, so 0 = 31^2 + 2(-6)d = 961 - 12d => d = 80 m

d. The velocity v(t) = d/t

is correct, because the velocity at time t can solved by dividing the distance traveled over the time taken

But the answer key says c. is correct, why is d. not also correct? Is it because dividing d/t would only give the AVERAGE velocity, not the velocity at time t?

Thanks,
Coop
 
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Coop said:
Is it because dividing d/t would only give the AVERAGE velocity, not the velocity at time t?
Exactly.
 
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Thanks :)
 
c cannot be correct unless you did dot copy it exactly as in the book.

d IS the stopping distance. So c will read
d=961-12*d
which is not true, is it?
 
Yes, nasu is correct. Choice c makes no sense. (I didn't read it before, I'm afraid.) Did you copy it correctly?
 
It's from a practice test made by my prof., directly choice c reads "The equation for the stopping distance is 961 - 12d = 0"
 
Coop said:
It's from a practice test made by my prof., directly choice c reads "The equation for the stopping distance is 961 - 12d = 0"
That's more like it. (Quite different from what you wrote the first time.)
 
Oh my mistake, I didn't realize, but now that you guys explained it I can see the difference
 

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