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Car Accelerating

  1. Sep 2, 2014 #1
    Hello all, here is my problem:

    A car accelerates, starting from rest, with a speed that is given by:

    v(t)=vm(1-e-at)

    a) What is the top speed of the car? Explain why.

    b) How far does the car travel in time t?

    c) Suppose the car can accelerate from 0 to 60 mph in 2.9s, and has a top speed of 195 mph. Imagine a mile-long race between two of these cars, with the same finish line, but with a different starting line: One drives along the ground towards the starting line at a point one-mile away, while the other is dropped out of a plane one mile above the ground. Which one reaches the finish line first? (ignore any air resistance for the falling car).

    That is the entire problem. I just need help getting started.

    First, what does the vm represent in the speed equation?
    Second, how exactly do I calculate the top speed?

    I also suppose that I will need the position and acceleration equations which I derived from speed equation for b and c respectively.

    I got:

    x(t)=vm*((e-at)/a+t)
    v(t)=vm*(1-e-at)
    a(t)=vm*(ae-at)
     
  2. jcsd
  3. Sep 2, 2014 #2

    Doc Al

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    Think of it as a constant. Once you answer part a, the notation will make more sense.

    You won't be able to get a numerical value, if that's what's throwing you off. Answer in terms of the constants given.

    You are given the speed as a function of time. If you graphed it, what would that function look like? What's its maximum value?
     
  4. Sep 2, 2014 #3
    I know I won't get a numerical value as the top speed. I just do not know what path to take to manipulate v(t) to get top speed.
     
  5. Sep 2, 2014 #4

    Doc Al

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    Does the speed increase or decrease as time increases?

    What's the maximum value of 1-e-at ?
     
  6. Sep 2, 2014 #5
    Increase in time=Increase in speed.

    Max value of 1-e-at = 0
     
  7. Sep 2, 2014 #6

    Doc Al

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    Good!

    Really? So the car doesn't even move? (Think that one over. :wink:)
     
  8. Sep 2, 2014 #7

    CWatters

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    Are you sure?
     
  9. Sep 2, 2014 #8
    Nevermind. Duh. I was thinking something completely different. So what is the a in the exponent? Acceleration due to gravity? If so then the max value would be 1, but that doesn't make sense.
     
  10. Sep 2, 2014 #9

    Doc Al

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    Treat 'a' as just another constant.

    Makes sense to me. At what time will the car have max speed? What is that max speed?
     
  11. Sep 2, 2014 #10
    So what I'm getting is that:

    v(t)=vm(1-e-at)

    We've said that the max value for (1-e-at) = 1

    So that means at max v(t)=vm

    In order to get that t=∞.
     
  12. Sep 2, 2014 #11

    Doc Al

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    Perfect! As time goes on, the speed gets closer to the maximum value of vm.

    Perhaps now you can guess what the 'm' stands for.
     
  13. Sep 2, 2014 #12
    I figured that's what it stood for I seem to have a habit of jumping to conclusions like that early on in the problem and confuse my self because I get determined to make that scenario be the case (right or wrong.)

    So part a is understood. Part b would be the distance equation setting t=∞?
     
  14. Sep 2, 2014 #13

    Doc Al

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    They don't want the distance at infinite time, but at time "t". (You've already solved that one in your first post.)
     
  15. Sep 2, 2014 #14
    See what I mean? Get ahead of myself and flustered and try to make a certain scenario fit. Then for part c I am given acceleration, the time for that acceleration, and top speed. Plug and chug then. Very helpful Al.
     
  16. Sep 2, 2014 #15

    Doc Al

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    Correction: You've almost solved it. Don't forget about the constant of integration.
     
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