1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    ! 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


    User Avatar
    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 !!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Problem related power Date
Solution of basic special relativity problem Feb 26, 2018
Galilean Relativity Problem Feb 6, 2018
Small Relativity Problem Jan 26, 2018
Invariant mass problem, elastic collision Jan 20, 2018
Work and Power related Problem Oct 25, 2015