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Homework Help: Problem related to power of engine and resistance

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A car of mass 500 kg has a maximum speed of 40 ms-1 on a level road with the engine of the car working at a constant power of 20 kW. The resistance to motion is proportional to the square of the speed.

    (i) find its acceleration when the speed is 20 ms-1
    (ii) find the distance traveled while the speed increases from 20 ms-1 to 30 ms-1
    (iii) when the car reaches the speed of 30 ms-1, the power is switched off. Find the time required to reduce the speed from 30 ms-1 to 20 ms-1

    2. Relevant equations
    P = Fv
    F = ma

    3. The attempt at a solution
    F = P/v = 20000/20 = 1000 N

    a = F/m = 1000/500 = 2 ms-2

    For (ii) and (iii), I don't know...
    the resistance = kv2, how to find the value of k? and how to continue after that? I think kinematics doesn't work because the acceleration is not constant.

  2. jcsd
  3. Feb 2, 2010 #2


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    ! the acceleration while a car is drivin at 20 m/s is 0. I think the problem meant the resistance or the force that the engine has to use.

    !! using simple x=Vo*t+1/2at^2 is easy but a = acc of the car - rezistance (which is a variable) Have fun!!!

    !!! again v=a*t where a=-rezistance (variable acording to speed) Have fun again!!! (Btw i'd say t=Infinity, just a wild guess that might be wrong)
  4. Feb 2, 2010 #3


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    Homework Helper

    Hi songoku,
    The net force acting on the car is given by
    f = F - k*v^2..........(1)
    When the car reaches its maximum velocity, its acceleration is zero. Hence f = 0.
    and F = k*v^2. But F = W/v. So
    W/v = k*v^2 or k = W/(vmax)^3. Find k.
    Now for (i), find f for given velocity by using eq(1) and hence find acceleration.
    For (ii), find f1 and f2 for two given velocities. Then
    ΔKE = (f1+f2)/2*s. find s.
    For (iii), put W = 0. m*a = -k*v^2 or a = -k/m*v^2
    So dv/dt = -(k/m)*v^2 Or
    dv/v^2 = -(k/m)*dt. Find the integration to get time t.
  5. Feb 3, 2010 #4
    H rl.bhat and Lok

    I think I get it. I'll try it first. Thanks !!
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