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Please explain this Kinematic Problem.

  1. Feb 11, 2012 #1
    How far do you travel if, starting from rest, you accelerate at 1.5 m/s2 for
    4.0 s, then coast for 8.0 seconds, then decelerate to a standstill in 4.0 s?

    I am totally confused about the fact you accelerate and then coast, and then slow down. We have 3-t's in this problem and don't know what formula to use.
  2. jcsd
  3. Feb 11, 2012 #2


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    Gold Member

    you probably want to divide the problem into these three stages. I find V-T diagrams helpful in these problems
  4. Feb 11, 2012 #3
    What are V-T diagrams?
  5. Feb 11, 2012 #4
    Think of it like this:
    The first part, you are accelerating at 1.5m/s (per second), therefore, after 4s you will be at v=??. Then the next part, you are coasting, therefore, not accelerating.
  6. Feb 11, 2012 #5
    So basically, I should just brake the question up into 3 steps?
  7. Feb 11, 2012 #6


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    sorry I mean velocity vs. Time diagrams, with area under the graph representing your displacement. You will need to use kinematics equation to find your peak velocity
  8. Feb 11, 2012 #7
    OK. I will try to work this out the way you told me to.
  9. Feb 11, 2012 #8
    Just look at it as three separate problems. First you travel distance when accelerating (constant acceleration,) then when you coast (constant velocity,) and then when you decelerate.

    The first you are given that you start from rest and accelerate at 1.5m/s^2 for 4 seconds. Using the derived expression for displacement in terms of what you are given:
    ΔD = v_i*t + 0.5*a*t^2 (v_i = velocity initial = 0) = 0.5*1.5m/s^2*(4s)^2 = 12m

    Next you are given that you coast for 8.0s. This is simply given by:
    ΔD = v_f*t (v_f = velocity final of last part)

    Where v_f = v_i + at = 0 + 1.5m/s^2*4.0s = 6.0m/s

    ΔD = 6.0m/s*8.0s = 48m

    For the last part you are given that you come to a stop in 4.0s. A short cut here is to pickout that this is the same amount of time that was taken to reach v_f. With constant deceleration, then, you know the same distance is taken as was when starting out, 12m.

    But to find it anyway, you have current velocity as 6.0m/s. The acceleration (in other direction) required for that to reach zero in 4.0s can be taken from v_f = v_i + at where v_f = 0m/s, v_i = 6.0m/s, and t = 4.0s. So 0 = 6.0m/s + a*4.0s => a = -1.5m/s^2. Which is expected as it is the same in magnitude as the beginning acceleration. So to get displacement:
    ΔD = v_i*t + 0.5at^2 = 6m/s*4.0s - 0.5*1.5m/s^2*(4s) = 12m

    Which is as said, the same as the displacement for part one. So total displacement is:
    D = 12m + 48m + 12m = 72m
  10. Feb 11, 2012 #9
    thank you sefrez.
    Last edited: Feb 11, 2012
  11. Feb 11, 2012 #10
    I did that because at first you are accelerating for 4 seconds. The next part you are coasting for 8 seconds. Your velocity is constant here.

    Edit, I see you edited your post. You must have figured out. If I had waited a few more seconds lol.
  12. Feb 11, 2012 #11
    Yes... lol i figured it out :).
    You can be sure you will be seeing me very often on this forum.
  13. Feb 11, 2012 #12


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    V-T graphs represent, what historians would call, the "primary source", (the original information) from which the 5 equations of motion are derived. Well four are derived from a V-T graph and the other is a mathematical manipulation of a couple of them.
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