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Homework Help: Accleration problem

  1. Sep 12, 2003 #1
    First off, thanks to all who helped me on my last questions. It is greatly appreciated that you are willing to help alleviate my physics stress! I am sure there will be many more questions to come. In fact, here is another physic's predicament that I am stranded over.

    Background: I am doing a project dealing with a car (Jaguar S-Type R). I have calculated the power (P = E/t) and the Kinetic energy (KE = 1/2mv2)

    Question: If the engine generates a constant power of Pavg, calculate the time it would take to accelerate from 0-30mph, 0-40mph, etc.

    I know the speeds (v0 = 0mph, v = 30mph), mass (1847.48kg), power (171.3078Hp, or 94219.284 ft-lbs/s), and Kinetic energy (664,537.1725J)

    So, can I calculate the Force by converting 94219.284ft-lbs/s into Newtons? I really do not know what ft-lbs/s can be converted. If I was able to determine the force of this car than I could use Newton’s second law equation: acceleration (Sum of Force)/(time) this would determine the acceleration; and then I could determine the sec by using this equation: a = change in v/ time

    I think I am completely off track, if any one could help me it would be greatly appreciated!!!
  2. jcsd
  3. Sep 12, 2003 #2


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    Staff Emeritus

    Power = work / time = W/t

    You state the power is equal to 171.3078Hp, which is equal to 127.744kW

    v = 30mph = 13.4112m/s

    K = (1/2)*m*v^2 = .5 * 1847.48 * 13.4112^2 = 166,144J

    Using the work-energy theorem W = K2 - K1, where K2 = 166,144J and K1 = 0J.

    W = 166,144J

    Therefore the time it takes to reach the power (or go from 0 to 30mph) is:

    t = W/P = 166,144J / 127.744kW = 1.3s
  4. Sep 12, 2003 #3


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    Science Advisor

    Notice that dduardo did NOT calculate force!

    Since power (i.e. "horsepower") is work/time (energy/time), he
    calculated the change in kinetic energy between "0-30mph, 0-40mph, etc." and then calculated how much time was required to give that much energy. In other words, he used "conservation of energy".
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