Car acceleration kinematics give average power

  • Thread starter late347
  • Start date
  • #1
301
15

Homework Statement



The performance of a car is being tested. Distance driven is 400m. The car accelerates from a standing start (v0 = 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

Ekin = (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 EKin. = W

The motionless car at V0, has infact also EKin. 0 = 0 Joules

EKin. 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 travelled 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)
 

Answers and Replies

  • #2
311
23
I sneaked a look at the answer, it looks as though air drag has been ignored, and average power has been calculated using the KE change and elapsed time.
 
  • #3
301
15
do you think I made erroneous assumption regarding Fs=W

I assumed F* 400m = 907 165 joules

and indede it does look like air drag was ignored along with other things like friction between wheel and ground.

It does look upon closer examination that the car would reach Vmax already before 400m. This could be done assuming acceleration as constant, and using
s= v0t + 0,5* at^2

However it is doubtful if this models real life car accelerations from standstill.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,598
5,991
do you think I made erroneous assumption regarding Fs=W

I assumed F* 400m = 907 165 joules

and indede it does look like air drag was ignored along with other things like friction between wheel and ground.
Yes, your calculation of the force was wrong. Even if the power were constant (which we are not told) the force would vary. You found the average force, but that does not help in finding average power. Dividing the work done by the time is the valid way to find that.

With regard to friction, it certainly does not ignore friction between tyre and ground. With no friction, the car would not move. You perhaps mean rolling resistance.
In fact, if you read the question carefully, you will find that it does not ignore drag or rolling resistance. It asks for the average power of the net force, i.e. the propulsive force remaining after subtracting drag etc.
 
  • #5
301
15
Yes, your calculation of the force was wrong. Even if the power were constant (which we are not told) the force would vary. You found the average force, but that does not help in finding average power. Dividing the work done by the time is the valid way to find that.

With regard to friction, it certainly does not ignore friction between tyre and ground. With no friction, the car would not move. You perhaps mean rolling resistance.
In fact, if you read the question carefully, you will find that it does not ignore drag or rolling resistance. It asks for the average power of the net force, i.e. the propulsive force remaining after subtracting drag etc.

No values were given for any frictions or air drag, though. I suppose one has to be content that that the average power was calculated.

It looks like the average power was to be calculated as the work done divided by time. I checked it from textbook again to be sure.

I suppose that in real life the friction would be varied in the beginning of the car journey. Because the car starts from a standstill, and should be accelerated in optimal manner , to its max velocity. Although 143km/h sounds quite fast speed indeed. Especially for regular car.

The performance tester driver ought to avoid "skidding the tyres", when accelerating from standstill etc...

Indeed this comes back to the idea that possibly less power is used, in the beginning acceleration portion, when accelerating from standstill.
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,598
5,991
I suppose that in real life the friction would be varied in the beginning of the car journey. Because the car starts from a standstill, and should be accelerated in optimal manner , to its max velocity. Although 143km/h sounds quite fast speed indeed. Especially for regular car.

The performance tester driver ought to avoid "skidding the tyres", when accelerating from standstill etc...

Indeed this comes back to the idea that possibly less power is used, in the beginning acceleration portion, when accelerating from standstill.
You don't seem to have comprehended the last para in my post. The question asks for the average power provided by the net force. That is, the remaining force after allowing for all losses due to drag, rolling resistance, skidding, whatever. The net force is what leads directly to the acceleration (ΣF=ma). Thus the answer is the gain in KE divided by the elapsed time. No assumptions, no approximations.
 

Related Threads on Car acceleration kinematics give average power

Replies
2
Views
2K
Replies
6
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
3
Views
4K
  • Last Post
Replies
7
Views
9K
Replies
4
Views
8K
Replies
4
Views
2K
Replies
8
Views
5K
Top