Does Faster Mean More Energy: Comparing 0-60 in 3 & 8 secs

In summary: And since Work = Force * distance, Power = Force * distance/time = Force * velocity. So more powerful engines can get you to the same speed faster, or a higher speed in the same time. So if you want more power, you need a bigger engine and more fuel consumption.So to get better gas mileage, you need a smaller engine (less power) or a more efficient engine (higher power for the same fuel consumption). And to get lower emissions, you need a more efficient engine (the power needed to overcome air resistance and go fast comes from burning fuel, and the less fuel you need to burn, the lower the emissions).In summary, when comparing two cars that accelerate from 0 to 60 in
  • #1
brycenrg
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Homework Statement


What takes more energy a car going 0 to 60 in 3 seconds or a car going 0 to 60 in 8 seconds.

Homework Equations


1/2 k v ^2

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?
 
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  • #2
brycenrg said:

Homework Statement


What takes more energy a car going 0 to 60 in 3 seconds or a car going 0 to 60 in 8 seconds.

Homework Equations


1/2 k v ^2

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

Can you write out all of the Relevant equations involved in this question? If you do that, you may see the answer...
 
  • #3
Sorry. 1/2 m v^2 - 1/2 mv^2
 
  • #4
brycenrg said:
Sorry. 1/2 m v^2 - 1/2 mv^2
That would equal zero... Equations have an equal sign between the lefthand side (LHS) and RHS... :smile:
 
  • #5
Lol it's w = 1/2 mv^2 - 1/2 mv^2
 
  • #6
brycenrg said:
Lol it's w = 1/2 mv^2 - 1/2 mv^2
So w = 0. o0)
 
  • #7
But it takes you more gas, a faster car uses more gas which is energy right?
 
  • #8
brycenrg said:
But it takes you more gas, a faster car uses more gas which is energy right?
I don't know offhand. I would have to look at the equations for the energies involved in this problem...
 
  • #9
Just talking about real life. I made up the problem
 
  • #10
brycenrg said:
Just talking about real life. I made up the problem
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?
 
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  • #11
berkeman said:
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?
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
brycenrg said:
What takes more energy a car going 0 to 60 in 3 seconds or a car going 0 to 60 in 8 seconds.
brycenrg said:
Just talking about real life. I made up the problem
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.
 
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  • #13
Thank you, that makes perfect sense. So the reason we have difference fuel economy for cars is because air resistance.
 
  • #14
brycenrg said:
So the reason we have difference fuel economy for cars is because air resistance.
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
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 doesn't 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
brycenrg said:
I just learned that it doesn't matter how fast you accelerate to the destination you will use the same amount of energy.
No, the question you asked was accelerating to the same speed, not the same distance... :smile:
brycenrg said:
But In my car if I punch the accelerator I get less mpg.
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
BTW, the equation for power is P = W/t, so doing the same amount of work faster requires more power.
 
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FAQ: Does Faster Mean More Energy: Comparing 0-60 in 3 & 8 secs

Does a car with a faster 0-60 time have more energy?

No, a car's 0-60 time does not directly correlate with its energy output. The 0-60 time is simply a measure of how quickly a car can accelerate from 0 to 60 miles per hour. Other factors such as engine power, weight, and aerodynamics also play a role in a car's energy efficiency.

Can a car with a slower 0-60 time still have more energy than a faster car?

Yes, a car with a slower 0-60 time can still have more energy than a faster car. This is because energy is not solely determined by acceleration, but also by factors such as torque and horsepower. A car with a slower 0-60 time may have a more powerful engine, resulting in higher energy output.

How does a car's 0-60 time affect its energy efficiency?

A car's 0-60 time does not have a direct impact on its energy efficiency. However, a faster 0-60 time may indicate that the car has a more efficient engine and can convert fuel into energy more efficiently. Additionally, a car with a slower 0-60 time may require more energy to maintain high speeds, resulting in lower energy efficiency.

Is a car with a faster 0-60 time always better?

No, a car with a faster 0-60 time is not always better. While a faster acceleration may be desirable for some drivers, it is not the only factor to consider when evaluating a car's performance. Other factors such as handling, braking, and fuel efficiency also play a significant role in overall performance.

Can a car's 0-60 time be improved to increase its energy?

No, a car's 0-60 time cannot be directly improved to increase its energy output. However, improving the car's engine and other components can result in a faster 0-60 time and potentially increase its overall energy efficiency. It is important to note that this improvement may also result in a higher energy demand, depending on the specific modifications made to the car.

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