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A toy car has a mass of 120g, including two 1.5V batteries connected in series.

  1. Sep 22, 2007 #1
    A toy car has a mass of 120g, including two 1.5V batteries connected in series. Suppose the motor that drives the car is 80% efficient (i.e., 80% of the electric energy goes to the drive wheels, 20% is dissipated as heat) and that friction and air resistance are negligible. How much charge passes through the batteries during the time it takes the car to accelerate from rest to 1.5 m/s?

    I thought i might be using these equations somehow. Not entirely sure though.

    P=IE
    I=[tex]\Delta[/tex]Q/[tex]\Delta[/tex]t ?

    I'm really having a hard time knowing where to begin. Any hints or thought provoking questions would help. I suppose i need to find out the amount of time it takes for the car to accelerate to that velocity first but i really have no idea how to begin.
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2
    What type of energy the car (including battery) is gaining ?
    What's the relation between that type of energy and the required speed (1.5 m/s) ?
    Figure this out and you might question the need of knowing the time that it takes.

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
  4. Sep 22, 2007 #3
    its gaining kinetic energy. KE= 1/2mv^2 i have a mass of 120g Velocity of 1.5m/s ... so

    KE = 1/2 (0.12Kg)(1.5)^2 = 0.27

    sorry had my decimal in the wrong place.. its 0.12kg not 1.2kg.
     
    Last edited: Sep 22, 2007
  5. Sep 22, 2007 #4

    mgb_phys

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    And the electrical power, P = current * volts. Energy is power * time and charge is current * time.
    So electrical energy E = I * V * t and Q = I * t
     
  6. Sep 22, 2007 #5
    careful, it's 120g not 1.2kg.

    nice, now what's the relation between that KE (at that speed) and the total charge that passed through the battery ?

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
  7. Sep 22, 2007 #6
    well knowing that E=IVt and Q=IT i can substitute Q for IT in the equation E=ItV and have E=QV
    i have a E of .27 and a total voltage of 3.0V? (two 1.5V batteries) .27/3=Q= 0.09 C?
     
  8. Sep 22, 2007 #7
    looks to me that you forgot the "1/2".
    The rest of calculations is right.

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
  9. Sep 22, 2007 #8
    how very right you are :). Thank you. It's always the simple things. so my Ke= .135 then and so .135/3=Q= 0.045 C? Would that be correct?
     
  10. Sep 22, 2007 #9
  11. Sep 22, 2007 #10
    Ok. Thank you guys.
     
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