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In which case car must do more work -- Kinematics

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    In which case car engine must do more work: To start driving from 0 kmph until car reaches 27 kmph or to reach 54 kmph from 27 kmph? Force of resistance in both cases are equal.

    I need to base answer of this problem on formulas. I apologize for any grammar mistakes (i'm not native english speaker).
    2. Relevant equations

    NA
    3. The attempt at a solution
    V0 = 0 kmph
    27 kmph = 7,5 mps
    54 kmph = 15 mps
     
  2. jcsd
  3. Jan 25, 2015 #2

    Stephen Tashi

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    What are the relevant equations for kinetic energies ?
     
  4. Jan 25, 2015 #3
    I think it's those equations:
    1) Fs = mdv / dt
    2) v = ds / dt
    but I have no clue how to use them.
     
  5. Jan 25, 2015 #4

    Stephen Tashi

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    Have you studied kinetic energy?
     
  6. Jan 25, 2015 #5
    Nope, never really
     
  7. Jan 25, 2015 #6

    Stephen Tashi

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    Let's make sure. The kinetic energy [itex] E [/itex] of a mass [itex] m [/itex] moving with velocity [itex] v [/itex] is [itex] E = \frac{1}{2} m v^2 [/itex].
     
  8. Jan 25, 2015 #7
    Mass isn't given but velocity is going to be higher to reach 57 kmph from 27 kmph, right?
     
  9. Jan 25, 2015 #8

    Stephen Tashi

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    Since the mass of the car stays the same, just represent it by a letter M.

    What do you mean "velocity is going to be higher"?
    Do you mean the "change in velocity"?
    No, the change in velocity from 0 to 27 is the same as the change in velocity from 27 to 57.

    But what is the change in kinetic energy from 0 to 27 versus the change in kinetic energy between 27 to 54 ?
     
  10. Jan 25, 2015 #9
    I mean, isn't velocity going to be higher when driving at 54 kmph?
    So, if the velocity doesn't changes (value stays the same) and E=½mv2 is the equation, then, change in kinetic energy also is the same?
     
  11. Jan 25, 2015 #10

    Stephen Tashi

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    Do you mean that velocity is higher at 54 kmph than at 27 kmph? Yes, that is correct. In English is you say "higher" , you need to compare two things. It doesn't make sense to say that 54 is "higher" unless you tell about another thing..

    Same as what? It doesn't make sense to say that a single thing is "the same". You have to compare two things.
     
  12. Jan 25, 2015 #11
    Yes, exactly

    But now it has confirmed that velocity will be higher while accelerating from 27 to 54 versus 0 to 27, right? So, if I put that unknown velocity value in equation as v2, change in kinetic energy (E) from 27 to 54 is also going to be higher versus 0 to 27?
     
  13. Jan 25, 2015 #12

    Stephen Tashi

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    The velocity values are not "unknown".

    If you compute the kinetic energy at the three known velocities, 0, 27, and 54, you can compute the change in kinetic energy beween 0 and 27 and the change in kinetic energy between 27 and 54.
     
  14. Jan 25, 2015 #13
    Why don't you do this:
    Consider the three states- 1) 0 kmph 2) 27 kmph 3) 54 kmph
    Find the the KE at each of these states. Then find the difference between states 2 &1 AND states 3 & 2.
    What do you find?
     
  15. Jan 25, 2015 #14
    Oh Stephen and me put up the same point I guess.
     
  16. Jan 25, 2015 #15
    Yup, i just made all calculations. (I used meters per second instead of km's per hour). Im not going to write it down, but I calculated that accelerating from 27 kmph to 54 kmph will use more KE versus 0 to 27. So, if we equalize KE to work, than car must do more work going from 27 to 54, right?
     
  17. Jan 26, 2015 #16

    Stephen Tashi

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    Correct.
     
  18. Jan 26, 2015 #17
  19. Jan 26, 2015 #18
    Thanks a lot!
     
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