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Acceleration from Kinetic energy.

  1. Sep 17, 2012 #1
    Hi,
    The question reads;

    A car motor produces 20Kw of power. All of which is transfered to the wheels producing drive, there is no resistance. the mass is 800kg. Calculate the acceleration after 7 seconds.

    I can see that it requires using dv/dt = a and rearanging Ke=1/2 mv^2 to include acceleration but when I tried using. v=u+at to form Ke= 0.5 M (AT)^2 I got an incorrect answer.

    HELP!
     
  2. jcsd
  3. Sep 17, 2012 #2
    This is not right because acceleration is not constant. Do you remember what the definition of "power" is? Start from that: assuming that power and mass are constants, you should be able to solve the speed of the car as a function of time.
     
  4. Sep 17, 2012 #3
    I knew it was wrong because of that, it was getting the velocity as a function of time which I was struggling with ?

    Thanks for looking!
     
  5. Sep 18, 2012 #4
    Bump! Still stuck??
     
  6. Sep 18, 2012 #5

    Doc Al

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    Staff: Mentor

    Your idea of taking a derivative is good, but what will you take the derivative of? What's the definition of power?
     
  7. Sep 19, 2012 #6
    Well I thought that power is rate of change of Ke in this circumstance, so d(Ke)/dt , but after that I got stuck. I thought if I could differentiate velocity with respect to time it may help?
     
  8. Sep 19, 2012 #7
    So if P = dKe/dt, then Ke = ∫Pdt. Solve this for velocity
     
  9. Sep 19, 2012 #8

    Doc Al

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    Staff: Mentor

    That's the right idea. Show what you got when you took the derivative.
    Well, sure. dv/dt = a, which is what you are trying to solve for.
     
  10. Sep 19, 2012 #9
    so If I differentiate both sides with respect to time:

    d(Ke)/dt = 0.5M(V^2) dv/dt

    I would get P = MA ?

    But that is incorrect as A varies?
     
  11. Sep 19, 2012 #10
    Where p is Power, M mass and a acceleration?
     
  12. Sep 19, 2012 #11
    That's not right. Remember chain rule! Now you have v=v(t), and mass is constant. What is d/dt v(t)2?

    I dont understand where you got that from.
     
  13. Sep 19, 2012 #12

    Doc Al

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    Staff: Mentor

    That's not quite right. Remember the chain rule.
     
  14. Sep 19, 2012 #13

    Curious3141

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    Homework Helper

    There is no need to use calculus here. An application of Power = Force * velocity and another of F = ma will allow expression of Power in terms of mass, (instantaneous) acceleration and (instantaneous) velocity. Express v in terms of the other variables.

    Then use ½mv^2 = KE = power*time. Put the previously derived expression for v here and solve for a in terms of power, mass and time.
     
  15. Sep 19, 2012 #14
    So using the chain rule would you treat V like a trig function? Then when differentiating with respect to time would you get,

    P = MAV ?

    Where P is d(Ke)/dt so Power

    M is the Mass which is constant

    A is the Acceleration

    V is the velocity.

    That gets the correct answer but is it the correct way? If it is thank you all so much for the help. It seems really simple now!
     
  16. Sep 19, 2012 #15

    Doc Al

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    You'd treat v as a function of t. d(KE)/dt = d(KE)/dv*dv/dt.

    Exactly.
     
  17. Sep 19, 2012 #16
    Thank you so much! I'm not quite sure why I had the block on treating V as a function. It seems so simple now!
     
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