Calculating Power Output of a Car Accelerating from 0 to 27 m/s in 60 seconds

[Medgirl314]

Homework Statement

A 1200 kg car accelerates from 0 to 27 m/s in 60 seconds.
a) What is the power outage from the engine in watts?
b) What is the power outage from the engine in horsepower?

Homework Equations

1/2 mgy=v
1 horsepower=746 Watts

The Attempt at a Solution

My final answer was 2025 Watts, 2.71 horsepower. Either I am way off, or my physics teacher is testing my confidence with physics. I'm guessing the former.

a) v=1/2(1200kg)(4.5m/s)²=12150 Joules=2025 Watts.
2025 Watts/746=2.71 horsepower.

Where did I go so terribly wrong??

Thanks!

 

[etudiant]

Is there not a slipped decimal?
Accelerating to 27m/sec in 60 seconds translates to .45m/sec acceleration.
You're showing that a car is not very efficient.

Separately, there may be some confusion. You should check your formulas.
What are the parameters V=1/2mgy? They do not clearly translate to v=1/2*1200kg*(4.5)**2

 

[Chestermiller]

Where did the 4.5 come from. It should be 27. So, in joules, what's the kinetic energy after 60 sec.?

Chet

 

[voko]

27 m/s is 60 mph or about 100 km/h. These days, cars typically go 0 - 60 mph in under ten seconds; the car in question takes 60 seconds. So just by that measure alone it packs very few ponies under its hood.

There was a car - a very popular one - Citroën 2CV, whose initial model only produced 9 hp. It needed over 40 seconds to get to 40 mph and more than eternity to reach 60 mph. Still, it was a practical car.

The very first car made by Karl Benz delivered less than 1 hp.

 

[AlephZero]

The question is badly worded IMO. It should say the average power outage.

If the acceleration is constant, the power will vary over the 60 seconds, and if the power is constant, the acceleration will vary. Constant power is probably a better approximation to real life.

As Chestermiller said, the way to do this is using energy. I don't recognize your formula 1/2 mgy=v but I don't think it is relevant anyway.

The answer I got was small, but not as small as the OP's. The answer will be less than in real life because it ignores the work done to overcome air resistance.

There was a car - a very popular one - Citroën 2CV, whose initial model only produced 9 hp. It needed over 40 seconds to get to 40 mph and more than eternity to reach 60 mph. Still, it was a practical car.

And they were as much fun to drive at 30 mph as most modern cars are at 90 ...
http://www.bruckmann.com/cars/2cv/2cv_13.jpg

 

[berkeman]

The question is badly worded IMO. It should say the average power outage.

If the acceleration is constant, the power will vary over the 60 seconds, and if the power is constant, the acceleration will vary. Constant power is probably a better approximation to real life.

As Chestermiller said, the way to do this is using energy. I don't recognize your formula 1/2 mgy=v but I don't think it is relevant anyway.

The answer I got was small, but not as small as the OP's. The answer will be less than in real life because it ignores the work done to overcome air resistance.



And they were as much fun to drive at 30 mph as most modern cars are at 90 ...
http://www.bruckmann.com/cars/2cv/2cv_13.jpg

Yoiks! That was obviously before anti-sway bars were invented...

 

[Medgirl314]

Is there not a slipped decimal?
Accelerating to 27m/sec in 60 seconds translates to .45m/sec acceleration.
You're showing that a car is not very efficient.

Separately, there may be some confusion. You should check your formulas.
What are the parameters V=1/2mgy? They do not clearly translate to v=1/2*1200kg*(4.5)**2

etudiant, yes on the slipped decimal. Sorry about the confusion, I typed 60 seconds, but I missed a decimal! It is actually 6.0 seconds.
The formula: m=mass,g=gravity,y=distance.WHICH I think is nonexistent. I combined PE=1/2mgy and KE=1/2mv². (Yay, I invented an equation! )
Oops?

Chet, I was trying to find m/s. The problem gives m/6 s. Sorry about the confusion, I typed 60 seconds, but I missed a decimal! It is actually 6.0 seconds. Okay, maybe I'll just get it now that I found those errors:
KE=1/2mv²
KE=1/2(1200)(4.5m/s² )
KE=1/2(1200)(20.25)
KE=12150 Joules
P=energy/time
P=2025 Watts
1 hp=746 W
2025*1 hp/746 Watts=2.71 horsepower.


Anndd no. I used the right formula on my paper but I typed it wrong here. So where did I go wrong? I sorted out the problems with the slipped decimal and the wrong formula. Is there some basal step I'm missing?

Thanks again, everyone!

 

[Chestermiller]

You still calculated the kinetic energy incorrectly for the end of 6 seconds. It should be:

(0.5)(1200)(27)² Joules. The average power (in watts) is this value divided by 6 seconds.

chet

 

[Medgirl314]

Oh, I missed that it's asking for the TOTAL power output, even though it was implied in some of your guys' posts. I just overcomplicated it.

(0.5)(1200)(27)2
=(0.5)(1200)(729)
=437400 Joules.
=437400/6 Watts
=72900 Watts
=72900/746 horsepower
=97.72 horsepower.

Seems much more likely to me. Could someone please check it?

 

[Chestermiller]

Oh, I missed that it's asking for the TOTAL power output, even though it was implied in some of your guys' posts. I just overcomplicated it.

(0.5)(1200)(27)2
=(0.5)(1200)(729)
=437400 Joules.
=437400/6 Watts
=72900 Watts
=72900/746 horsepower
=97.72 horsepower.

Seems much more likely to me. Could someone please check it?

Looks good.

Chet

 

[Medgirl314]

Thank you!

 

[Mark44]

Slightly off-topic, but the 2CV model name stood for deux chevaux vitesse, which is the speed of two horses, if my French translation is at all accurate. I never understood how two horses could be faster than one. Maybe it meant that if you wanted the car to go really fast, get two horses to pull it.

 

[voko]

Slightly off-topic, but the 2CV model name stood for deux chevaux vitesse, which is the speed of two horses, if my French translation is at all accurate.

That corresponds to "deux chevaux-vapeur", which literally means "two steam horses", but in fact means "two tax horsepowers", which is basically a taxation class based in the displacement of the engine, which was close to two real horsepowers in the beginning of the 20th century, but as time went on and the engine efficiencies increased, the discrepancy became very significant.

I never understood how two horses could be faster than one.

This, BTW, is widely held misconception. In just about any western, a horseback rider can easily catch up with and outrun a multi-horse carriage. While in reality it is not necessarily so.

 

[Medgirl314]

I suppose it would depend on the mass and design of the carriage,correct? As well as the exact energy of the horses?

 

[voko]

I suppose it would depend on the mass and design of the carriage,correct?

Yes. Its payload, too.

As well as the exact energy of the horses?

The power developed by the horses is the more accurate way of saying that.

 

[Medgirl314]

The power developed by the horses is the more accurate way of saying that.

Thank you! I couldn't figure out how to put it.