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Acceleration and velocity of a car

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data


    The 400 kg mine car is hoisted up the incline with a slope
    angle of 30°. The massless cable connects the mine car and
    the motor that induces its motion. The force in the cable is
    defined by following function F = 3200 t2 [N], where the
    time t is in seconds. The mine car has at t = 0 and s = 0 an
    initial velocity v1 = 2 m/s. There is no friction between the
    mine car and the incline.
    a) Determine the acceleration of the mine car for t = 1 s.
    b) Calculate the velocity of the mine car when t = 2 s.

    Image of the mine car
    http://img14.imageshack.us/i/caraws.jpg/" [Broken]


    http://img14.imageshack.us/i/caraws.jpg

    2. Relevant equations

    F=ma

    mgsin(30) + 3200 t^2 = ma - second derivative - here we have acceleration

    now we integrate the above and we get

    a= (mgsin(30) + 3200t^2)t / m - first derivative - here we have velocity

    now we integrate again and we get

    a= (mgsin(30) + 3200t^2)t^2 / 2m - here we have the distance

    3. The attempt at a solution

    Are my equations correct ?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 16, 2009 #2
    No.

    In the first equation the sign of the force due to gravity and the tension in the cable are the same.

    Also you neglect the initial velocity of 2m/s.
     
  4. Dec 17, 2009 #3
    Ok.Right. Here is what i do now.

    F=m*a

    F-mgsin(30) = m*a

    3200t2 - 400*9.8*1/2 = 400*a

    a=8t2 -4.9 - thats the acceleration

    We need to find the acceleration after 1 second. Therefore a=8-4.9=3.1 -> a=3.1

    Now we have to find the velocity after 2 seconds.

    a=dv/dt -> dv=a*dt . Now we integrate and we get the velocity after 2 seconds.


    Are my equations now correct ?
     
  5. Dec 17, 2009 #4
    Looks good, but possibly neglects the initial velocity which needs to be entered. Perhaps you planned on adding the constant after the integration which is fine, just wanted to make sure it wasn't forgetten: ie [tex]\int dv[/tex]=V(t) +Vo
     
  6. Dec 17, 2009 #5
    Yes. Thank you. And one last question. I need to determine the position of the car after 3 seconds. Can I use this formula -> S=S0 + v0t + 1/2at2
     
  7. Dec 17, 2009 #6
    No that is used for linear acceleration; instead integrate again, So = 0 in the problem.
     
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