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Motion Questions -- Acceleration of a race car...

  1. Feb 10, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    I have been struggling to understand how to do this question so some guidance would be great.
    A racing car of mass 941kg accelerates from 28km/h to 102km/h over a distance of 297m. Frictional forces and wind resistance can be assumed to be 1160N.
    Determine the following:
    a) The average acceleration
    b) The time taken to accelerate from 28km/h to 102km/h
    c)The tractive force produced by the car to provide this acceleration
    d)The car finally reaches a speed of 212km/h. Friction and wind resistance are 2761 at this speed. What power output is required to maintain this speed?

    I have been playing around with equations and numbers and have just got myself really confused now, any help would be greatly appreciated.

    This is my attempt:
     

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    Last edited: Feb 10, 2016
  2. jcsd
  3. Feb 10, 2016 #2

    SteamKing

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    Show us your best effort so far.
     
  4. Feb 10, 2016 #3
    ok i will send a picture
     

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  5. Feb 10, 2016 #4

    haruspex

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    I could follow your working down to the point where you calculated time. All that looks ok.
    You then calculated a force, but have not stated what that force is, so I cannot say if you have that right. How are you defining F there?

    At this point I would like to grasp the problem setter by the collar and scream into his/her face FRICTION IS WHAT MAKES THE CAR ACCELERATE. Without friction the wheels would just spin. Presumably the problem setter meant air resistance plus rolling resistance equals 1160N.
     
  6. Feb 10, 2016 #5
    Yes it was part c and D that got me really confused any advice on how to answer that? I just guessed and plugged in the mass 941 and the acceleration 1.25 into f=ma to get that number.
     
  7. Feb 10, 2016 #6

    haruspex

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    You still have not defined that f. How is f defined in f=ma?
     
  8. Feb 10, 2016 #7
    I think i am defining it as the amount of force required to make the car accelerate at the value 1.25m/s^2. But is that tractive force then?
     
  9. Feb 10, 2016 #8

    haruspex

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    In the equation F=ma, F is the net force. What forces act to produce this net force?
     
  10. Feb 10, 2016 #9
    The resistant force 1160 so it would be 941 * 1.25 and then that subtract that resistant force to get the actual driving force?
     
  11. Feb 10, 2016 #10

    haruspex

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    Write an equation relating Fnet, Fresist, Fdriving. You can either take the same direction as positive for each, so some may turn out negative in value, or take each of those values to be positive and so define different directions for them. Your choice, but say which you choose and be consistent.
     
  12. Feb 10, 2016 #11
    aah so if Fnet is Fresist + Fdriving then Fdriving is mass * acceleration - (-Fresist) since Fresist is in the opposite direction
     
  13. Feb 10, 2016 #12

    haruspex

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    Yes.
     
  14. Mar 25, 2017 #13
    I am going through a similar example question. My question is in regards to part d.) The car reaches a (v) of 212km/h and maintains this (v) thus the acceleration would be (0m/s^3) so that would mean the tractive force aka applied force (Fa) would be lower with the lack of a net force that is just a curiosity not so much part of the question.
    So the formula for the power output is P= E used/time taken or P=FV in this situation there are two forces the tractive force and the friction force both working against the mass.
    I feel like im thinking wrong about the tractive force as it states the tractive force is what provides the acceleration.
    Any help much appreciated I have been scrambling my head on the different forces for awhile and need someone to help me put them in the right order. cheers.
     
  15. Mar 25, 2017 #14

    haruspex

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    Not sure what you mean by that. Are you misusing the term "friction force" to mean a force opposing forward motion, in the same way as the problem setter (see my post #4), or are you using it correctly? If you mean the force opposing motion, they work against each other.
    The net force, tractive force minus opposing forces (drag+rolling resistance), provides the acceleration.
     
  16. Mar 26, 2017 #15
    I would think the wind and air resistance (Ff2) would be opposing forward motion?
    So to work out net force the formula is Fnet=ma which would be 0N as there is no acceleration?

    This is the formula I have for working out tractive force where Force applied(Fa) is the tractive force:
    Fa = Fn + Ff
    Fa = 0 + 2761
    Fa = 2761N
     

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  17. Mar 26, 2017 #16

    haruspex

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    Wind and air resistance are the same thing. The other force opposing motion is rolling resistance. This consists of several components: axle friction, road surface flexion and tyre flexion.
    Your diagram is wrong. The tractive force acts in the forward direction (or the car would not go forward).
    Your choice of subscripts for the forces is not helpful. "a" suggests acceleration, "n" suggests normal and "f" suggests friction.
    I'll use t for traction, r for resistance and net for net.
    By definition, the net force is the sum of all the applied forces. That is a vectorial statement, but applies to scalars too as long as we are careful with the signs; i.e., choose a positive direction for all forces and the acceleration.
    Fnet=Ft+Fr.
    In your diagram, if we choose left to right as positive then Ft is positive and Fr is negative.
    Since there is no acceleration, Fnet=0.

    See also section 4 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/
     
  18. Mar 26, 2017 #17
    That clears up the subscripts big time cheers. If Fnet = 0N and Fr = -2761N due to the direction (right to left) then Ft = +2761N.

    So the formula for power is E out/time taken or P=FV
    So we use the Traction force to work out power as it is the means to the car moving?
    P=2761x58.89=162595.29J = 162.6KJ = 162.6KW
     
  19. Mar 26, 2017 #18

    haruspex

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    Yes, but try to handle the units consistently.
    P = 2761N x 58.89m/s = 162595 Nm/s = 162595W = 162.6kW.
    Not J, KJ, or KW.
     
  20. Mar 26, 2017 #19
    I think Im getting mixed up with energy and I have no excuse for the KW lol. Cheers
     
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