Non-constant acceleration of a race car from rest

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Homework Help Overview

The problem involves a race car starting from rest and attempting to cover a distance of 440 yards in the shortest time possible, with a focus on understanding the nature of its acceleration. The context includes calculations related to average acceleration, constant acceleration, and the implications of non-constant acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of equations for average and constant acceleration, questioning their applicability in the context of non-constant acceleration. There is an exploration of how to logically explain the non-constant nature of acceleration based on the provided data.

Discussion Status

Participants are actively engaging with the problem, offering insights and corrections regarding the equations used. Some express uncertainty about how to approach the problem given the non-constant acceleration, while others suggest that contradictions arising from applying constant acceleration equations could serve as proof of non-constant acceleration.

Contextual Notes

There is an acknowledgment that the problem is situated within a calculus-based course, which may influence the methods and reasoning employed by participants. The initial information provided includes specific values for distance, time, and final speed, which are central to the discussion.

haroldwershow
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Homework Statement



"In a drag race a car starts at rest and attempts to cover 440 yd in the shortest possible time. The world record is 5.637s; the final speed was 250.69 mi/h at the 440 yd mark"

a)What was the average acceleration?
b)Prove that the car did not move with constant acceleration
c)What would have been the final speed if the car had moved with constant acceleration so as to reach 440 yd in 5.637 s?

V_initial: 0 ft/s V_final: 367.68 ft/s T: 5.637 s Distance: 1320 ft

Homework Equations



For a) avg a = ΔV / Δt
for b) constant a = (V_initial + V_final) / 2
for c) V_final = V_initial + a_constant * time



The Attempt at a Solution



a) Avg a = 65.23 ft/s^2 No problems here

b) Using the constant acceleration equation above (which I am somewhat dubious of), a_constant = 183.84 ft / s^2. Intuitively, I believe that the average acceleration would have to be equal to the constant acceleration if this was indeed a case of constant acceleration. I'm not quite sure i can defend that logically.

c) Using the constant acceleration calculated above, V_final = 1036 ft/s, which is clearly way too high (V_final was given as 367.68 ft/s). So I think perhaps this high velocity answer proves part b.

How do I logically explain part b? Am I correct in using the constant acceleration equation described above? Or is there a totally different way to address this problem?

Thank you!

 
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I am pretty sure I am correct in saying that the equations you state,
For a) avg a = ΔV / Δt
for b) constant a = (V_initial + V_final) / 2
for c) V_final = V_initial + a_constant * time
are only valid if the acceleration is constant.
If the acceleration is not constant,. then this becomes a calculus problem.
 
haroldwershow said:
For a) avg a = ΔV / Δt
Yes, that's valid regardless of whether accn is constant.
for b) constant a = (V_initial + V_final) / 2
That can't be what you mean. (It's dimensionally wrong for a start.) Maybe you meant: for constant a, avg v = (V_initial + V_final) / 2. That is true, but not the most helpful. Do you know an equation, valid for constant accn, that relates accn, distance, and initial and final speeds?
 
haroldwershow said:
b) V_average = (V_initial + V_final) / 2
c) V_final = V_initial + a * Δt
Combine b and c:

V_average = (V_initial + (V_initial + a Δt) / 2 = V_initial + 1/2 a Δt

then distance:

distance = V_average Δt = V_initial Δt + 1/2 a Δt^2

for this problem V_initial = 0.
 
Thanks for the responses!

#2 The problem statement seems to imply that the acceleration is not constant. This is why I am finding it tricky; all the equations I know are valid for constant acceleration. This is a calculus based course. I'm familiar with the basic derivative relationships between distance, velocity and acceleration. How might I use them for this problem?

#3 You are right, " for b) constant a = (V_initial + V_final) / 2 " is dimensionally incorrect, it should be Avg V = (V_initial + V_final) / 2. This appears to only be valid for constant acceleration. Which brings me back to my original problem - this is a case of non-constant acceleration. So the constant acceleration questions appear to be invalid.

#4 I am given distance in the initial information. I am trying to prove that the acceleration is not constant.

Any ideas?

Thanks!
 
haroldwershow said:
Thanks for the responses!

#2 The problem statement seems to imply that the acceleration is not constant. This is why I am finding it tricky; all the equations I know are valid for constant acceleration. This is a calculus based course. I'm familiar with the basic derivative relationships between distance, velocity and acceleration. How might I use them for this problem?

#3 You are right, " for b) constant a = (V_initial + V_final) / 2 " is dimensionally incorrect, it should be Avg V = (V_initial + V_final) / 2. This appears to only be valid for constant acceleration. Which brings me back to my original problem - this is a case of non-constant acceleration. So the constant acceleration questions appear to be invalid.

#4 I am given distance in the initial information. I am trying to prove that the acceleration is not constant.

Any ideas?

Thanks!
You seem to be missing the key piece of logic. If you apply formulae which are valid when acceleration is constant and you arrive at a contradiction in the provided data, then you will have proved acceleration is not constant.
So I ask again: Do you know an equation, valid for constant accn, that relates accn, distance, and initial and final speeds?
 

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