Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Quick physics question

Options
  • 07-05-2007 4:44pm
    #1
    Registered Users Posts: 929 ✭✭✭


    Bit stuck, its prob very easy but cant seem to be able to start it...

    A car of mass 1200kg which is travelling at 100km/h wants to stop in a distance of 100m to avoid a collision. If the maximum retarding force due to breaks is 2000N, will the car avoid the collision? What is the minimum force needed to avoid collision?

    Cheers!!


Comments

  • Registered Users Posts: 5,851 ✭✭✭PurpleFistMixer


    This is quite possibly wrong but I'll give it a go anyway...

    You want to find the acceleration (or deceleration as the case may be) of the car, so...
    v^2 = u^2 + 2as...
    100km/h = 1000/36 m/s = u
    v = 0
    a = ?
    s = 100

    So a comes at as.. -3.86 or so.

    Then, F=ma, want to see the force required to cause the deceleration, so
    a = 3.86
    m = 1200
    F therefore = 4629.6N
    therefore, retarding force due to breaks isn't sufficient, and the F there is the minimum force needed.

    I think!


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Wrong.

    The answer was "lol, retarding".


    You can also just say 2Fs/m must be greater than or equal to u^2 for the breaks to stop the car, but 2(2000)(100)/1200 < (1000/36)^2, therefore they won't, no need to get the accelleration. For the minimum force to stop the car, F = m(u^2)/2s


  • Closed Accounts Posts: 51 ✭✭ron-burgandy


    JC 2K3 wrote:
    Wrong.

    The answer was "lol, retarding".


    You can also just say 2Fs/m must be greater than or equal to u^2 for the breaks to stop the car, but 2(2000)(100)/1200 < (1000/36)^2, therefore they won't, no need to get the acceleration. For the minimum force to stop the car, F = m(u^2)/2s

    Could you tell me what you derived that from?


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Well for the minimum force required to stop the car:
    v^2 = u^2 + 2as
    v = 0
    a = F/m
    0 = u^2 + 2Fs/m
    u^2 = -2Fs/m
    However, since the force is backwards you can write it as u^2 = 2Fs/m (you can leave it as it was above, but remember you'll have to take the 2000N as -2000N since it's in the opposite direction to the velocity vector)


Advertisement