POWER - different approach, different results

Basically, I tried to find the solution by calculating P=Fs/t, where F= 2250 and s is the distance traveled in the 12th second and that result differs from the result I get when I calculate the power using P=Fv.

##F=ma=1500*1.5=2250N##
##s_{12}-s_{11}= \frac{1}{2}at_{12}^2-\frac{1}{2}at_{11}^2=(0,5*1,5*144)-(0,5*1,5*121)=108-90,75=17.25m##
##P=\frac{Fs}{t}=\frac{2250*17,25}{1}=38812,5W##

##v_{12}=a*t=1,5*12=18##
##P=F*v_{12}=40500W##

Can somebody please tell me where I am wrong? Thank you.

Which is the instantaneous and which is the average power?
You want power averaged over time. This is given by $$\bar P=\frac{\int_0^tP~dt}{\Delta t}=\frac{\int_0^tF~v~~dt}{\Delta t}=\frac{\int_0^t (ma)(at)~dt}{\Delta t}.$$
Power units are Watts, not Newtons.

kuruman said:

Which is the instantaneous and which is the average power?
You want power averaged over time. This is given by $$\bar P=\frac{\int_0^tP~dt}{\Delta t}=\frac{\int_0^t (ma)(at)~dt}{\Delta t}.$$
Power units are Watts, not Newtons.

No, I want instantaneous power in 12th second. I didn't solve for average power here, I just typed the entire problem. I know how to solve for average, so that is not important. But I do not know how to solve for instantaneous power. I didn't learn calculus so I don't know what the symbols you wrote that mean.

Yes, Watts, my bad. Thanks.

Edit: I don't know why I typed the entire problem, I edited it.

Like, I do not know calculus, so I can't understand the reasoning behind it. So I am confused, which method shoud I use for instantaneous power P=Fv or P=Fs/t?

Iamconfused123 said:

Like, I do not know calculus, so I can't understand the reasoning behind it. So I am confused, which method shoud I use for instantaneous power P=Fv or P=Fs/t?

You should use P=Fv but note that v ≠ s/t because the car moves under constant acceleration and v = at.

If you don't know calculus, in this case you can use ##\bar P = F\bar v##, i.e. the average power over the time interval is equal to the force times the average velocity over that interval.

kuruman said:

You should use P=Fv but note that v ≠ s/t because the car moves under constant acceleration and v = at.

Thanks, but can you explain to me why this formula works and the other one does not? But without calculus, just words.

Also, do you recommend I learn calculus first because I am getting really frustrated by having to guess which formula works and which one does not. And if so, up to what level of calculus should I learn 1, 2, 3?
How long would that take? I just covered trigonometry (addition, double and half angles), would I be able to do it without much knowledge of trig? like, I haven't covered grahphs, trig. equations, trig. inequalities, etc.

Iamconfused123 said:

Homework Statement: ... Calculate average Power and instantaneous power in the 12th second.

There is a mistake in the question which makes it unanswerable.

The 12th second runs from t=11s to t=12s. But the term ‘instantaneous power’ applies to a particular value of t. Without this value the question can't be answered.

I'd guess the intended question is to find the instantaneous power at t=12s (assuming that there is no discontinuity in motion at t=12s, i.e. that the car's acceleration is constant either side of t=12s).

If my guess is correct, you need to find the velocity at t=12s, then you can use P = Fv.

Steve4Physics said:

f my guess is correct, you need to find the velocity at t=12s, then you can use P = Fv.

I assumed that your guess is correct. I read "in the 12th second" to mean "at the 12th second."

Steve4Physics said:

There is a mistake in the question which makes it unanswerable.

The 12th second runs from t=11s to t=12s. But the term ‘instantaneous power’ applies to a particular value of t. Without this value the question can't be answered.

I'd guess the intended question is to find the instantaneous power at t=12s (assuming that there is no discontinuity in motion at t=12s, i.e. that the car's acceleration is constant either side of t=12s).

If my guess is correct, you need to find the velocity at t=12s, then you can use P = Fv.

yeah, but like, distance at the end of the 12th second or velocity at the end of 12th second, does it matter? if I have the formula for both and both should give back the same result?

PeroK said:

Distance travelled has nothing to do with power.

In my textbook, one formula for power is P=W/t, where W=Fs and "s" is the path on which force is acting up on the object. In this case wouldn't that be the distance crossed? Assuming no friction, drag or losses of energy

Iamconfused123 said:

Thanks, but can you explain to me why this formula works and the other one does not? But without calculus, just words.

The only formula to use for instantaneous power is ##P=Fv## where ##v## is the instantaneous velocity. The equation ##v=s/t## is used when the acceleration is zero and the object covers equal distances in equal times. This is not the case here because the acceleration is not zero which means that the object does not cover equal distances in equal times. The instantaneous velocity is not constant but changes according to ##v=at## which is what you should use instead of ##s/t##. So here ##P=Fv=(ma)(at)=ma^2t## gives the instantaneous power.

kuruman said:

The only formula to use for instantaneous power is ##P=Fv## where ##v## is the instantaneous velocity. The equation ##v=s/t## is used when the acceleration is zero and the object covers equal distances in equal times. This is not the case here because the acceleration is not zero which means that the object does not cover equal distances in equal times. The instantaneous velocity is not constant but changes according to ##v=at## which is what you should use instead of ##s/t##. So here ##P=Fv=(ma)(at)=ma^2t## gives the instantaneous power.

Got it now. Thank you very very much.

Iamconfused123 said:

In my textbook, one formula for power is P=W/t, where W=Fs and "s" is the path on which force is acting up on the object. In this case wouldn't that be the distance crossed? Assuming no friction, drag or losses of energy

I deleted that post when I realised it wouldn't help with your confusion over average and instantaneous quantities.

kuruman said:

I assumed that your guess is correct. I read "in the 12th second" to mean "at the 12th second."

Yes. In case what I said was misconstrued, note that Post #7 was intended only for the OP. I thought that they may have been confused by the poor wording in the question.

It did not relate to anything you had written - apologies if it appeared otherwise.

Steve4Physics said:

Yes. In case what I said was misconstrued, note that Post #7 was intended only for the OP. I thought that they may have been confused by the poor wording in the question.

It did not relate to anything you had written - apologies if it appeared otherwise.

Not to worry. Nothing was misconstrued.

kuruman said:

I assumed that your guess is correct. I read "in the 12th second" to mean "at the 12th second."

That doesn’t help me. At the start of the 12th second or the end?

@Iamconfused123, is the statement in post #1 exactly as given to you? If not, please post the original wording (or maybe it is a translation?)
The intent could be:

1. at t=12s
2. at t=11s
3. the average over the 12th second

Your second attempt correctly solved (1).
Your first attempt found (3).

haruspex said:

That doesn’t help me. At the start of the 12th second or the end?

At ##t = 12~##s substituted in ##P=ma^2t^2##. Is that at the start of the 12th second or the end?

The original statement of the problem (now edited) also asked for the average power which I assumed to be from 0 to 12 s. That wording is lost.

kuruman said:

At t=12 s substituted in P=ma2t2. Is that at the start of the 12th second or the end?

That is not the same as "at the twelfth second". The 12th second lasts from t=11s to t=12s.

Sure. The problem says "A car with a mass of 1500 kg accelerates at 1.5m/s^2 from rest, and for 12 seconds reaches a velocity of 18m/s." To me "for 12 seconds" means that 12 seconds have elapsed and ##t=12~##s in the equation.

kuruman said:

Sure. The problem says "A car with a mass of 1500 kg accelerates at 1.5m/s^2 from rest, and for 12 seconds reaches a velocity of 18m/s." To me "for 12 seconds" means that 12 seconds have elapsed and ##t=12~##s in the equation.

But the question is asked in the next sentence. It could have been asked as "at t=11s" without creating any contradiction, so that interpretation is not ruled out.

haruspex said:

But the question is asked in the next sentence. It could have been asked as "at t=11s" without creating any contradiction, so that interpretation is not ruled out.

Please see post #8 for my interpretation of the question asked. OP is from Germany and the formulation of the question may be a translation.

To @Iamconfused123: Can you post the statement of the problem exactly as was given to you?

kuruman said:

Please see post #8 for my interpretation of the question asked.

which, as I wrote in post #16, I can interpret as either t=11s or t=12s.
Now, I do agree that t=12s is the most likely intent. It just is not certain.

haruspex said:

That doesn’t help me. At the start of the 12th second or the end?

@Iamconfused123, is the statement in post #1 exactly as given to you? If not, please post the original wording (or maybe it is a translation?)
The intent could be:

1. at t=12s
2. at t=11s
3. the average over the 12th second

Your second attempt correctly solved (1).
Your first attempt found (3).

It is exactly as posted. Although it is a translation, "in" which I posted refers to the period from 11th to 12th second. We do not have the word "at" in Croatian so it can be confusing sometimes, but if that was the case I would expect the author to write for "at the 12th second" something like "at the end of 11th second".
The author should have been clearer about what she meant.
It can be frustrating, but the collection has 5000 problems for high-school physics, every high school uses it and it's relatively good, so yeah.
@kuruman

Iamconfused123 said:

It is exactly as posted. Although it is a translation, "in" which I posted refers to the period from 11th to 12th second. We do not have the word "at" in Croatian so it can be confusing sometimes, but if that was the case I would expect the author to write for "at the 12th second" something like "at the end of 11th second".
The author should have been clearer about what she meant.
It can be frustrating, but the collection has 5000 problems for high-school physics, every high school uses it and it's relatively good, so yeah.
@kuruman

Perhaps you could find a physics instructor who understands Croatian, show this person the various interpretations mentioned in this thread and ask which one is appropriate to this problem. I think that you now understand the basic physics behind this problem. If you are using it for practice, there is probably nothing more to learn from it except for how to translate Croatian problems into English with minimum loss in translation.