- #1

late347

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## Homework Statement

The performance of a car is being tested. Distance driven is 400m. The car accelerates from a standing start (v

_{0}= 0).

(reader's note: It could be plausibly assumed that car performance tests are measured at even ground.)

The car's mass is 1150kg. The time taken in the acceleration is 16,1 seconds. The car's final velocity is 143km/h

Calculate the average power of the total force (net force?), which accelerates the car.

## Homework Equations

delta E = W

P= W/ t

Fs= W

E

_{kin}= (1/2) * (m) * (v^2)

= 56 kW power

## The Attempt at a Solution

We assume even ground at the car testing event. Likewise standing start means that the car starts from standing place motionless.

I had difficulty with this problem because I got the wrong result. Evidently there was a mistake or oversight in my own assumptions, upon which I had based my calculation.

It is known that potential energy does not change in this situation, with respect to the road. The car stays upon the road, and does not elevate itself with respect to the road at least greatly so.We need to transform km/h into m/s. 143km/h = 39,72 m/s

v0= 0

v1= 39,72 m/s

The work done is calculated using kinetic energies...

delta E

_{Kin.}= W

The motionless car at V

_{0}, has infact also E

_{Kin. 0}= 0 Joules

E

_{Kin. 1}= 0,5 * 1150kg * (39,72^2)

= work done

= 907165,08 Joules. This is the maximum kinetic energy of the car, I think.

Then I calculated the force. Fs= W. F= W/s.

907165,08 / 400m

F=2267,91 Newtons

I guess then I calculated P = (Fs)/ t

[400m * 2267,91 N ] / 16,1 seconds

__56 345,59 Watts__

I got a nagging feeling that I looked at the question wrongly at this point, even though the result looks roughly ok...Maybe I was overthinking it, and it could simply be calculated as

P= W/t

average power = (Work done by force) / (time used to do that work)

Is that the correct interpretation ? I looked back at a book example and that was the formula to be used for average power

Anyhow... This is inputted as.

907165,08 Joules / 16,1 seconds =

__56 345,65 Watts__roughly 56 kW power..

Why am I calculating only the average power, such that, I divide the work done, by the time taken during acceleration? I guess that's where I got confused.

Doesn't the acceleration force affect the car all the time? Why focus upon only the first 16,1 seconds?

I guess in other words, when we focus on the first 16,1 seconds, we are focusing upon the acceleration portion, of the entire journey of the car (entire journey was 400m)

So, in other words we would no longer be answering the question about the power of the accelerating force, if we were to examine all the portions of the car's journey? I.e. the entire journey 400m

It does appear that it is possible, that the car reaches its Vmax, already before the 400m mark at the ground. This would mean that the car drives the accelerating portion, and the final portion of the journey is traveled at constant speed Vmax essentially (Vmax= 39,72m/s)

In real life I think we could not know for sure, unless we knew exactly what the acceleration of the car was. Was it constant or not?

Average acceleration could be calculated as

a= (delta V) / (delta t)

39,72m/s / 16,1 seconds

average acceleration is 2,4670 m/s^2

Using that value it look like the car reaches Vmax already at the distance of 319,74m.

One could use s= (v0*t) + (0,5*a*t^2)