Power Calculation for a Car Accelerating to 87 km/hr in 6.3 s | 940 kg Mass

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Homework Help Overview

The discussion revolves around calculating the power developed by a car's engine as it accelerates from rest to 87 km/hr in 6.3 seconds, with a mass of 940 kg. The original poster attempts to solve the problem using constant acceleration but questions the implications of the problem stating "constant power."

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinetic energy and power concepts, with some exploring the relationship between energy and power over time. There are questions about the implications of constant power versus constant acceleration.

Discussion Status

Some participants have provided guidance on using energy concepts to approach the problem. There is an ongoing exploration of thermodynamic efficiency and its potential impact on the calculations. Multiple interpretations of the problem are being considered, particularly regarding the assumptions of constant power.

Contextual Notes

Participants note a lack of relevant examples in their textbook and mention previous lectures discussing thermodynamic efficiency, which may influence their understanding of the problem.

TraceBusta
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A car manufacturer claims that their car can accelerate from rest to 87 km/hr in 6.3 s. The car's mass is 940 kg. Assuming that this performance is achieved at constant power, determine the power developed by the car's engine.
Ok, I solved the problem by first solving for constant acceleration. a=[v(f)-v(0)]/t
Then using that I found the Force, using F=ma. Then plugging into P=F(dot)v

The answer that i got, 87143.056W is wrong. I think it could be because the problem says it is constant power, not constant acceleration.

How can I go about solving this problem? I don't really understand the concept of constant power instead of constant acceleration
 
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Remember what power is. The unit for power, watts, is equal to joules per second. Power is energy transferred in a given amount of time. Can you solve the problem using energy concepts?
 
ok, i know that Kinetic energy is = .5mv^2

so the kinetic energy at v=87 km/hr (24.167 m/s) is .5(940 kg)(24.167 m/s)^2= 11358.49 J. I don't know where to go from here.
I divided it by 6.3 sec to get 1802.93 J/s, but that is wrong. Atleast the units are right, heh.
 
I think you have the right idea and your numbers look right. Was there any discussion of thermodynamic efficiency in your class or your textbook?
 
in the last lecture the prof. talked about it for a few minutes at the end. my book has some examples, but nothing related to this problem, (or so i think)
 
TraceBusta said:
in the last lecture the prof. talked about it for a few minutes at the end. my book has some examples, but nothing related to this problem, (or so i think)

That may be a big clue! Just to get an idea is "the answer" approximately 3 times as large as the one you got?
 
well i won't know the right answer until the computer says "OK"after i enter it. until then it just says "NO"

I'm curious, why would it be 3 times larger?
 
TraceBusta said:
well i won't know the right answer until the computer says "OK"after i enter it. until then it just says "NO"

I'm curious, why would it be 3 times larger?

It's only a guess but thermodynamic efficiency is generally about 30%

If I might ask, how do you know the answer you got is wrong?
 
it says "no" when i put in a wrong answer
 
  • #10
TraceBusta said:
ok, i know that Kinetic energy is = .5mv^2

so the kinetic energy at v=87 km/hr (24.167 m/s) is .5(940 kg)(24.167 m/s)^2= 11358.49 J. .


24.167^2*940/2 = 274500 :smile:

ehild
 
  • #11
Problem

How would you find this same type of problem, but in addition, there's a constant air resistance force? Thanks.
 
  • #12
Subtract work done by frictional forces from the total energy. (W=FdcosΘ)
 

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