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

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The discussion centers on calculating the power required for a car with a mass of 940 kg to accelerate to 87 km/hr in 6.3 seconds, emphasizing the distinction between constant power and constant acceleration. Initial calculations using force and kinetic energy concepts led to an incorrect power output, prompting a deeper exploration of thermodynamic efficiency, which may suggest the actual power output is approximately three times the initial estimate. Participants noted the importance of considering energy loss due to factors like air resistance, which would require adjusting the total energy calculations. The conversation highlights the need for clarity on how constant power affects the dynamics of the problem. Ultimately, understanding these concepts is crucial for accurately solving similar physics problems.
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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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