# Non-constant acceleration of a race car from rest

1. Apr 24, 2013

### haroldwershow

1. The problem statement, all variables and given/known data

"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

2. Relevant 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

3. 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 24, 2013

### barryj

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.

3. Apr 24, 2013

### haruspex

Yes, that's valid regardless of whether accn is constant.
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?

4. Apr 24, 2013

### rcgldr

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.

5. Apr 26, 2013

### haroldwershow

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!

6. Apr 26, 2013

### haruspex

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?