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Kinematics - Car accelerating from traffic light

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    The spacing between two traffic lights is 300m. A car can accelerate from rest with a
    constant acceleration of 4 m/s2 and decelerate at 8 m/s2. (Ignore the reaction time of the
    deriver.)
    a) What is the maximum speed the car can reach and still be able to stop right at the
    second traffic lights when it starts from rest at the first traffic lights?
    b) Under these circumstances how long does it take for the car to travel between the two
    traffic lights?


    3. The attempt at a solution
    I know that displacement (x) = 300m
    Acceleration (a) = 4m/s2 and -8m/s2
    initial velocity (Vo) = 0 m/s
    -After this I am stumped on what equation to use. Are we trying to find time for both and final velocity for both and if so what equation do we use, thanks!
     
  2. jcsd
  3. Feb 10, 2010 #2
    Re: Kinematics

    The journey, which has a total distance of 300 metres, can be broken up into two parts:
    i) The accelerating part
    ii) The decelerating part
    During i), the driver accelerates at 4m/s^2 until he reaches a certain speed. In order for the speed to be maximum, part ii) must begin immediately upon reaching that speed; ie driver immediately decelerates at 8m/s^2 after reaching that certain speed.

    Thus, the way to do this question would be to split up the situation:
    [tex]v_{max}^2 = 0^2 + 2a_{1}s_{1}[/tex]

    [tex]0 = v_{max}^2 + 2a_{2}s_{2}[/tex]

    [tex]s_{1} + s_{2} = 300[/tex]

    Alternatively, simple observation by symmetry provides an easier way out.
     
  4. Feb 10, 2010 #3

    rl.bhat

    User Avatar
    Homework Helper

    Re: Kinematics

    Using kinematic equation Vmax^2 - Vo^2 = 2*a*x, you can show that
    a1*x1 = a2*x2.
    Or x2/x1 = a1/a2. Add 1 on both side. You get
    x2/x1 + 1 =a1/a2 + 1.
    x = x1 + x2 and it is given. Now solve for x1 and x2.
     
  5. Feb 10, 2010 #4
    Re: Kinematics

    so confused right now.

    Vmax^2=0^2+2a1s1 which would be

    Vmax^2 = 2(4)s1 = 8s1

    Then:
    0 = Vmax^2 + 2(a2s2) which would be

    Vmax^2 = -2(-8)s2 = 16s2

    am I on the right track so far or am I just screwing this up. I am studying for a test and just do not get this problem
     
  6. Feb 10, 2010 #5
    Re: Kinematics

    Yes, you are on the right track. Now, we simply equate the two expressions for Vmax^2 to obtain:
    [tex]8s_{1} = 16s_{2}[/tex]
    [tex]2s_{1} = s_{2}[/tex]

    Since we know that [tex]s_{1} + s_{2} = 300[/tex], we can easily get the values of s1 and s2, and hence [tex]v_{max}[/tex].
     
  7. Feb 10, 2010 #6
    Re: Kinematics

    ok so this is what I got for part a:
    8x1 = 16x2, so x1 = 2x2, so 2x2 + x2 = 300, which is x2 = 100, so x1=200
    then plug x1 into vmax equation which gives u 40m/s.

    For part B. do I use the equation x=.5at^2 + Vot, and do this for both x1 and x2 and then add the times together?
     
  8. Feb 11, 2010 #7
    Re: Kinematics

    Yes, that works - but leaves you with a quadratic equation for the decelerating part. It would be simpler to just use v = u + at to solve for the respective times.
     
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