Non-constant acceleration of a race car from rest

AI Thread Summary
In the discussion about a drag race car's acceleration, participants analyze the car's performance over 440 yards, noting that it starts from rest and achieves a final speed of 250.69 mi/h in 5.637 seconds. The average acceleration calculated is 65.23 ft/s², but participants question the validity of using constant acceleration equations, as they lead to contradictions with the given final speed. It is established that if applying constant acceleration formulas results in unrealistic values, it indicates that the acceleration is not constant throughout the race. The conversation emphasizes the need for calculus-based approaches to accurately describe the car's motion under non-constant acceleration conditions. Understanding these dynamics is crucial for solving the problem effectively.
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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