# Homework Help: Energy concept

1. Mar 4, 2016

### brycenrg

1. The problem statement, all variables and given/known data
What takes more energy a car going 0 to 60 in 3 seconds or a car going 0 to 60 in 8 seconds.

2. Relevant equations
1/2 k v ^2

3. The attempt at a solution
I know the answer is they are the same but the power is different. But doesn't the car that goes faster lose more gas. And then it would need more energy?

2. Mar 4, 2016

### Staff: Mentor

Your "Relevant Equation" is not an equation, and looks to be incorrect anyway...

Can you write out all of the Relevant equations involved in this question? If you do that, you may see the answer...

3. Mar 4, 2016

### brycenrg

Sorry. 1/2 m v^2 - 1/2 mv^2

4. Mar 4, 2016

### Staff: Mentor

That would equal zero... Equations have an equal sign between the lefthand side (LHS) and RHS...

5. Mar 4, 2016

### brycenrg

Lol it's w = 1/2 mv^2 - 1/2 mv^2

6. Mar 4, 2016

So w = 0.

7. Mar 4, 2016

### brycenrg

But it takes you more gas, a faster car uses more gas which is energy right?

8. Mar 4, 2016

### Staff: Mentor

I don't know offhand. I would have to look at the equations for the energies involved in this problem...

9. Mar 4, 2016

### brycenrg

10. Mar 4, 2016

### Staff: Mentor

In real life, you use equations to understand the situation.

What is the equation that relates kinetic energy KE to mass and velocity?

What is the equation that relates power P to KE?

What is the equation that relates Work done W to Power?

11. Mar 4, 2016

### brycenrg

The only equation I have learned so far is W = KEf-KEi
I guess there is other equations I need to know to understand the question

12. Mar 4, 2016

### Staff: Mentor

Since you just made up the problem to try to understand it better, I'll show you how to use equations to figure it out.

The equations you use are the kinematic equations for constant acceleration (assuming the cars have constant acceleration, and start from rest):

Velocity is acceleration multiplied by time: v(t) = a * t

Distance is related to the acceleration and time squared: d(t) = 1/2 a * t^2

Force is mass multiplied by acceleration: F = ma

And the Work done is the product of the force and the distance: W = F * d

Since both cars reach the final speed of 60m/h, we can use the velocity equation to find the ratio of their accelerations:

v(final) = 60 = a1 * t1 = a2 * t2

So a1/a2 = t2/t1 = 8/3

And we can take the ratio of the distances: [d1/d2] = [(1/2 a1` * t1^2) / (1/2 a2 * t2^2)] = [(a1 * 3^2) / (a2 * 8^2)] = (a1/a2) * (3^2/8^2) = (8/3) * (3/8)^2 = 3/8

Finally we can get the ratio of the forces F1/F2 = (m1 * a1)/(m2 * a2) = a1/a2 = 8/3 since the cars have equal mass.

So we can now calculate the ratio of the work done for the two accelerating cars:

W1/W2 = (F1 * d1) / (F2 * d2) = F1/F2 * d1/d2 = a1/a2 * d1/d2 = 8/3 * 3/8 = 1

So the quicker car accelerates over a shorter distance, but uses the same amount of energy as the car accelerating to the same speed over a longer distance (neglecting air resistance, etc.). Does that start to make some sense? And that's why the final Kinetic Energy KE for each is the same, since they have the same mass and the same velocity at the end.

13. Mar 4, 2016

### brycenrg

Thank you, that makes perfect sense. So the reason we have difference fuel economy for cars is because air resistance.

14. Mar 4, 2016

### Staff: Mentor

Not really. You would need to make your question more specific. If you are asking why a car that can go 0-60 in 3 seconds gets worse fuel economy than the car that can only do it in 8 seconds, the reason is it takes a more powerful engine to accelerate faster. And more powerful engines tend to get lower gas mileage.

15. Mar 4, 2016

### brycenrg

Yeah that was the question.
Is it possible for a more powerful engine to get higher gas mileage? Or if not why is that? I just learned that it doesnt matter how fast you accelerate to the destination you will use the same amount of energy. But In my car if I punch the accelerator I get less mpg. Just trying to wrap my head around it. Thanks for helping me.

16. Mar 4, 2016

### Staff: Mentor

No, the question you asked was accelerating to the same speed, not the same distance...
To get the higher power, you need to consume fuel at a faster rate, hence the lower gas mileage when you punch it.

Engines can have features that help to increase gas mileage at a given power (like turbochargers), but in general more power means more fuel consumption.

17. Mar 4, 2016

### Staff: Mentor

BTW, the equation for power is P = W/t, so doing the same amount of work faster requires more power.