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How to Find the Theoretical 0-60 mph Time of a Car?

  1. Sep 3, 2014 #1
    1. The problem statement, all variables and given/known data
    A car weighs 2200 lbs, has a peak torque of 750 lb ft at 6500 RPM. Assume a CVT and racing differential keeps the power to all four wheels of the car constant and equal. What is the theoretical 0 to 60 mph time?


    2. Relevant equations
    [itex] Fd = \frac{1}{2}m v_f ^2 - \frac{1}{2}m v_0 ^2 [/itex]
    [itex] P = \frac{\tau \times \omega}{\Delta t}[/itex]


    3. The attempt at a solution
    So the power output of the engine is going to be [itex] P = \frac{(750 lb ft)(6500 rpm)(2\pi rad)}{60 sec} = 510,250 \frac{ft lbs}{sec}[/itex]
    The amount of work needed to get the car up to 60 mph from rest ideally would be [itex] W = \frac{1}{2}(2200 lbs)(\frac{60 mi * 5280 \frac{ft}{mi}}{3600 sec})^2 = 8,518,400 ft lbs [/itex]
    Dividing those two numbers gives us 16.7 sec. I don't know if my approach is right with the work-energy theorem or if I have to use some other metric for calculating the time because I'm sure it's not that slow if the engine is pushing nearly 1000 hp...
     
  2. jcsd
  3. Sep 3, 2014 #2

    Matterwave

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    Science Advisor
    Gold Member

    Yeah, you confused the units a little bit. lbs is a unit of force, you have to convert that to a unit of mass for it to work in the work equation (divide by acceleration due to gravity here on Earth).

    You can see a problem occurs because your final units in the work equation are lbs*ft^2/s^2 and not lbs*ft
     
  4. Sep 3, 2014 #3
    But wouldn't that mean my units for work will be slug ft^2 / s^2? How would I convert that to compare it to the power of the engine?
     
  5. Sep 3, 2014 #4
    Wait, slug ft/s^2 = lb, so that would work... thanks
     
  6. Sep 3, 2014 #5

    rcgldr

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    Homework Helper

    Side note - normally a CVT allows an engine to operate at the rpm for peak power, not peak torque, since with optimal gearing, maximum torque at the real wheels for any given speed (except zero or near zero speeds in a real world situation) corresponds to maximum power from the engine.
     
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