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Please help, Resistive Forces, Initial Acceleration.

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The mass of a sports car is 900 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.250 and the frontal area is 2.00 m2. Neglecting all other sources of friction, calculate the initial acceleration of the car ,if it has been traveling at 100 km/h and is now shifted into neutral and is allowed to coast. (Take the density of air to be 1.295 kg/m2.)

    2. Relevant equations

    R=.5 *D*p*A*v^2

    3. The attempt at a solution

    I calculated it out and got that R is equal to 9, however I am not quite sure what to do with the Resistive Force.
     
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    LowlyPion

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    Doesn't F = m*a still?
     
  4. Feb 22, 2009 #3
    I guess it would, so does that mean that i would just do 9=ma?
    because i did that but then you get like .01 and thats not correct. Plus i know the answer has to be negative..
     
  5. Feb 22, 2009 #4

    LowlyPion

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    Well of course it's decelerating.

    Apply minus sign accordingly.
     
  6. Feb 22, 2009 #5

    LowlyPion

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    Btw, I didn't calculate your number. I suppose you converted your units correctly?
     
  7. Feb 22, 2009 #6
    i believe i did,
    the drag coeff. is still .25
    the frontal area is 2 m^2
    100 km/h converts to 27.8 m/s
    and i think the density of the air is still 1.295 km/m^2

    After i did that plugged them into the equation i got that the ressistance force is 17.99

    so 17.99=(900) (a)?

    a= .02

    Where did i go wrong?
     
  8. Feb 22, 2009 #7
    im sorry, density of air was kg/m^2
     
  9. Feb 22, 2009 #8

    LowlyPion

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    That would be different from your first answer of 9.

    And it is (-) because the force is resistive to the direction of motion.
     
  10. Feb 22, 2009 #9
    o im sorry, i forgot to plug 1/2 in the beginning of the equation, so it really is 9, and then that divided by 900 equals .01.

    but -.01 isnt the correct initial velocity..
     
  11. Feb 22, 2009 #10

    LowlyPion

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    You mean deceleration don't you?
     
  12. Feb 22, 2009 #11
    yes, thats what i meant, but do you have any idea what im doing wrong?
     
  13. Feb 22, 2009 #12

    LowlyPion

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    R = .5*(.25)(1.295)*(2)*(27.8)2

    and that only comes out to 9?

    I get a different answer.
     
  14. Feb 22, 2009 #13
    oo thank you, i seemed to forget to square the velocity for some reason?
    but anyways, thank you very much!
     
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