# Easy questions.

1. Apr 16, 2005

### whozum

Went to the tracks today:
Car weighs 3560 pounds, which is 1615kg.
I ran a quarter mile (402m) in 17.8s.

So naturally being the physics student I found the average acceleration to be

$$a = \frac{2d}{t^2} = \frac{804}{316.84} = 2.54m/s^2$$

Average force applied would then be F = m a = 1615kg * 2.54m/s^2 = 4102N

Work = 4102N * 402m = 1649044Nm

and the Power = 1649044Nm / 17.8s = 92642W = 124 horses.

What I dont get is my car is rated at 103 horses, how would I get 124 horses at the wheel? I think the error would be in the acceleration being the average acceleration, but wouldnt it come out to the same since average acceleration is pretty much "if you were accelerating at xxx m/s^2 you would have acheived the same distance in the same time". So it would pretty much be the same right?

2. Apr 16, 2005

### marlon

Why do you take two times the distance in your formula for a?

marlon

3. Apr 16, 2005

### ramollari

whozum

First, did you start measuring the time from the moment you started the car? And second, are you sure about the mass of your car?

marlon

d = at^2/2 -> a = 2d/t^2

4. Apr 16, 2005

### marlon

But that's not the average acceleration !!!
marlon

5. Apr 16, 2005

### shyboy

the power is determined as an energy change in a time interval divided by the time interval. The higher the engine power, the faster you can change the speed of your car (which I will NOT recomend to try) But you can estimate the limit and you will find out your personal contribution to the global warming (joke)

6. Apr 16, 2005

### whozum

What would it be?

7. Apr 16, 2005

### whozum

It was at a drag race track, and they start the clock as soon as your wheel gets off the starting line, and stops as soon as it touches the finish line. I re-looked up my car's weight and found it to have a 'curb weight' of 2500lbs, so I figured with me in it it would be another 160 for a total of 2660. This is 1210kg. I re-did the power output with the new mass (I initially had thought it was somewhere along the lines of 1600kg) and the power came out to 60-80hp, depending on the gear. (I redid the calculation and found the power output during certain intervals of the run).

I can post all the calculations if someone wants to look at them, but does it make sense to lose 40% power in the drivetrain?

Using the average acceleration times distance over time then should give me power no?

Apparently I was wrong in this assumption.

8. Apr 16, 2005

### whozum

My traptime was 76mph in 17.8sec, so in that case avg acceleration would be 4.26m/s^2. When I calculated power with this though it came out alot higher.

9. Apr 16, 2005

### shyboy

I did not quite understand your numbers, but let's do it anyway.

If you can can get 76 mph from the rest in 17.8 s, that meants that
you average acceleration was

$$a=\frac{76 mph}{17.8s}=\frac{76\times 1609 m}{3600s\times 17.8}=1.9 m/s^2$$

The average force is about 2300 N

The kinetic energy of the car is
$$E=mv^2/2=1210 kg*(76\times 1609 m/3600 s)^2/2=700 kJ$$
the average power is
$$P=E/t=40 kW=50 hp$$

Now, 124 hp seems to be the maximum engine power. If you are idling it should be much less. You also loose some power against air resistance, which is pretty strong at high speeds. The number at 76 mph is about

$$F_{air}\sim 0.5 S \rho_{air} v^2\sim 1400 N$$

S is an effective area, which is of the order of the car crossection.

The air resistance is a tricky thing, it strongly depend on the car shape.

10. Apr 16, 2005

### whozum

Hang on. Air resistance has already played its role in the acceleration. Why do our accelerations differ so much?

We are neglecting air resistance because our acceleration already accomodates air resistance. It is the net effective acceleration. I'm starting to really doubt these numbers if I'm putting out 50hp.

124hp was obtained the same way you did your measurements but it was using the wrong mass. The correct calculation I used $P = madt^-1$ with $m = 1210kg, a=2.54, d=402m, t=17.8s$ and got 93.17hp.

If we use the acceleration with a = v/t. 77mph = 0.02138mps = 34.2m/s. This gives 1.92m/s^2 like you said.

$P = madt^-1 = \frac{(1210kg)(1.9m/s^2)(402)}{17.8s} = 69.7hp$

I dont know about engine efficiencies and stuff but thats 70% of the engine power. I dontk now if thats alot more than I should expect to be lost in the drivetrain, but its reasonable.

The problem is finding the correct acceleration to use, and probably harder since it wasnt constant.

11. Apr 16, 2005

### shyboy

Well, you don't know how much did you lose against air. You can measure only the car speed, which is going against the air, so you engine is increasing the car speed AND pushing the air. Thus the resulting force acting on the car is less then the force supplied by the engine.
You are right the our accelaration accomodates the air resistance, but you are missing that in the absence of the outside air you speed will be larger if you will do the same racing. It looks like the main losses are due the air resistance, so drive slow, save the money ;) - or buy something sporty-racy. I belive, there is a term "crusing speed", which means that at this speed a vehile engine works the most effectively

If you take the average acceleration as a constant $$1.9 m/s^2$$, you will get the distance about 301 m. Which means your acceleartion was not uniform.
There is a definition of averaging: to find the average speed, you need to divide the way on the time. To find the average acceleration, you need to divide the final velocity on the time. To find the average power you need to divide the kinetic energy change on the time.

That is because when you do the averaging, you summarize values at different intervals and then divide on the number of intervals:

$$\overline{x}=\frac{\sum x_i} {N}=\frac{\sum x_i\delta t_i} {N \delta t_i}=\frac{\sum \delta y_i} {T}=\frac{\Delta y} {T}$$

here $$\delta y=x\delta t$$

Last edited: Apr 16, 2005
12. Apr 16, 2005

### whozum

This isall true but it isnt what I'm asking about. Im familiar with allthis.

I know the acceleration wasnt uniform, but I'm trying to approximate a value of uniform acceleration that would do the same distance in the same time.

As you said,$$a = v/t = 34.2/17.8 = 1.9m/s^2[/itex] [tex] x = \frac{(1.9)(17.8^2)}{2} = 300$$

So your method (though correct) doesnt give us the right answer, for some reason.

Rearranging the above gives:

$$a = 2dt^{-2} = 2(402)(\frac{1}{316.84}) = 2.54m/s^2[/itex] Last edited: Apr 16, 2005 13. Apr 16, 2005 ### shyboy You are trying to calculate the average acceleration from the distance and time. That is incorrect. It works only if the acceleration is constant. That is why 300m<>400m, the actual travel path. the correct average power accelerating the car is 50 hp if you start from zero and hit the finish at 76 mph. To find out how much you lose, you need to know what is the average engine power (it may not be 103 hp). When I mentioned 120 hp I thought that was the rated car's maximum power. BTW, at any given moment your power is [tex]P=mva$$

so the average power will be

$$P=m\overline{av}$$

you first mistake is

$$\overline{a}=\frac{2d}{t^2}=\frac{2 \overline{v}}{t}<>\frac{v_{max}}{t}$$

your second mistake is that you multiply the force averaged over time (if it was yor intention?) on the distance. You can multiply the averaged force on the distance only if it was averaged over distance. But it make little sense because what we want to find out the work, which is, we know, equals the car's kinetic energy.

Last edited: Apr 16, 2005
14. Apr 17, 2005

### whozum

First mistake: It was actually:

$$\overline a = \frac{\Delta v}{\Delta t}$$ which is a correct definition.

Second: This could get weird but I get what your saying. The rated power is meaningless because the power includes the power lost in the drivetrain as well as the power lost to air resistance, thus the power isnt really corresponding to the engine's ability. Right?

15. Apr 17, 2005

### shyboy

right, but the rated power is not meaningless. You will not say that the salary is meaningless because you need to pay taxes ;)

Last edited: Apr 17, 2005
16. Apr 17, 2005

### whozum

I mean the experiment doesnt represent the rated power, so it has no value on finding the ratedpower. I know what youmean i just suck at saying it.