Question: How can I calculate the power output of a car accelerating uphill?

[Carrie9]

Homework Statement

A car of 1200kg accelerates up a 10 degree incline and accelerates from 30 km/h to 90 km/h in 8 sec. What is the power needed

Relevant Equations

P=w/t

So my result is 102hp, but it’s not correct.

 

[PeroK]

Homework Statement:: A car of 1200kg accelerates up a 10 degree incline and accelerates from 30 km/h to 90 km/h in 8 sec. What is the power needed
Relevant Equations:: P=w/t

So my result is 102hp, but it’s not correct.

You need to show your working.

 

[Carrie9]

a=((25-8,3)/8)=2,1 m/s^2
Fg=1200(9,8)sin10=2042,1N

Fa=ma+Fg
Fa=1200(2,1)+2042,1=4562,1N

P=Fa v
P=Fa ((Vo+Vf)/2)
=75958,965 W

=102hp

 

[PeroK]

a=((25-8,3)/8)=2,1 m/s^2
Fg=1200(9,8)sin10=2042,1N

Fa=ma+Fg
Fa=1200(2,1)+2042,1=4562,1N

P=Fa v
P=Fa ((Vo+Vf)/2)
=75958,965 W

=102hp

If you have constant acceleration, do you have constant power?

Or, if you have constant power, do you have constant acceleration?

 

[Carrie9]

If you have constant acceleration, do you have constant power?

Or, if you have constant power, do you have constant acceleration?

I suppose yes

 

[PeroK]

I suppose yes

$P = Fv$

As the speed increases the force decreases under constant power.

 

[Carrie9]

Sorry but I don’t get it

 

[PeroK]

Sorry but I don’t get it

What don't you get? $F = P/v$.

If $P$ is contant and $v$ increases, then $F$ decreases.

That's why if you run, cycle or drive a car, you can't just just keep going faster and faster.

 

[Carrie9]

I don’t get how this changes my calculations

 

[PeroK]

I don’t get how this changes my calculations

It means that your calculations, which assumed constant acceleration and constant force, are wrong.

The clue is from your OP:

Relevant Equations:: P=w/t

 

[Carrie9]

Now I’m just confused. How do I calculate the W then? and where do I apply the forces

 

[PeroK]

Now I’m just confused. How do I calculate the W then? and where do I apply the forces

Forget forces. Power is about energy.

 

[Carrie9]

Forget forces. Power is about energy.

So I need to divide the change in kinetic energy by the time? KE=(1/2)mv^2 where m is the mass (1200kg) and v is the speed (25 and 8,3 m/s)?
am I completely lost?

 

[PeroK]

So I need to divide the change in kinetic energy by the time? KE=(1/2)mv^2 where m is the mass (1200kg) and v is the speed (25 and 8,3 m/s)?
am I completely lost?

Remember the car is also going uphill.

 

[Carrie9]

So change in KE + change in PE?
PE=mg(hf-hi) where hf-hi= s sin 10

 

[PeroK]

So change in KE + change in PE?
PE=mg(hf-hi) where hf-hi= s sin 10

Do you have any more information on this problem?

Either you can assume constant power; or you could assume constant acceleration and calculate the maximum power.

The second is simpler. You might try that first and see whether that gets the correct answer.

 

[Carrie9]

Do you have any more information on this problem?

Either you can assume constant power; or you could assume constant acceleration and calculate the maximum power.

No more info, got the answer 75,7W now

 

[PeroK]

No more info, got the answer 75,7W now

Okay, that's the average power. Hmm.

The power increases linearly with speed, so the maximum required is $113kW$.

 

[Carrie9]

Okay, that's the average power. Hmm.

The power increases linearly with speed, so the maximum required is $113kW$.

Thank you very much for the help

 

[gneill]

@Carrie9 It's great that you got to the solution!

Here's a short way to get the same result.

First, find the acceleration as you did, so:

$a = \frac{v_f - v_i}{\Delta t} = 2.083 \frac{m}{s^2}$

Now the force required to accelerate a car with acceleration $a$ on a slope of angle $\theta = 10°$ is:

$F = M \left(a + g \, \sin(\theta) \right)$

The power being delivered by the engine at the start and finish of the trip up the hill:

$P_i = F \, v_i = 37.86 \,kW$

$P_f = F \, v_f = 113.59 \,kW$