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Graphing X versus Time

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data

    At t= 0s a truck is stopped at a traffic light. When the light turns green, the truck starts to speed up, and gains speed at a constant rate until it reaches a speed of 20m/s 8 seconds after the light turns green. The truck continues at a constant speed for 60m . Then the truck driver sees a red light up ahead at the next intersection, and starts slowing down at a constant rate. The truck stops at the red light, 180m from where it was at .
    Draw accurate graph for the motion of the truck.
    2. Relevant equations
    x-x_0 = t.2(v initial + v final)

    3. The attempt at a solution

    I would think that the line is linear, since as time passes by the distance away from origing keeps on increasing.

    But the problem states that it speeds up until it reaches constant speed of 20m/s.
    So for the first 8 seconds, the speed will increase until it reaches 20m/s.
    That is the problem I have having in starting.

    I would use the equation listed above, but

    x final = ?
    x initital = 0
    t= 1 (this number varies, as I plot points throughout)
    v initial = 0
    v final = ? (I would not know how this would work, as I know that velocity would be 20m/s at 8seconds, before it must be increasing)
    and

    a = ? (unknown to me)

    How would I approach this problem.
    What I've done thusfar.
    [​IMG]
     
  2. jcsd
  3. Sep 14, 2007 #2

    CompuChip

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    The truck gains speed at a constant rate. This means that the velocity changes linearly. So in a (v, t) plot, you would indeed get straight lines. In particular, between 8 seconds and 60 m further, the velocity line would be horizontal and the motion line (which you wish to draw) is linear.
    Now during the speeding up and slowing down, the velocity changes at a constant rate. This rate of change is called the acceleration. If the acceleration is constant, the velocity after a time t is given by [itex]v(t) = a \times t[/itex]. It is related to the position through [itex]x(t) = \frac12 a t^2[/itex]. What does this tell you about the form of the graph? Can you find the acceleration and make it (quantitatively) precise?
     
  4. Sep 15, 2007 #3
    I am sorry, but I do not understand.

    But since you mentioned velocity vs time graph. So for the first 8second it will be linear from 0,0 to 8,20, after that it will be a horizontal line for the remaining time it's velocity is constant and then decreases toward the end. But now how do I find the time, it stays the same, and the time it decreases?

    Hopefully this will help me in the first graph.
     
  5. Sep 15, 2007 #4
    You can use speed=distance/time to find the time- so you can calculate how long he was going at a constant speed for (he went 20m/s for 60m) then he slowed down over 180m from a speed of 20m/s to 0m/s.

    Hope this makes sense!
     
  6. Sep 15, 2007 #5
    so what you are trying to say is that speed = 20m/s, is equal to distance (60) over time (unknown) = that would equal 3.

    20 = 60/t = 20t= 60 = t = 60/20 t= 3

    ???
     
  7. Sep 15, 2007 #6

    CompuChip

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    You don't really need the (v, t) graph, I just hoped you saw what you were doing. But clearly you're not, so let me try another approach :smile:

    Let me call the three phases in the motion 1, 2 and 3. Recall that with constant acceleration a (acceleration = rate of change of velocity), the traveled distance [itex]\Delta x[/itex] in a time [itex]\Delta t[/itex] is [itex]\Delta x = \tfrac12 a (\Delta t)^2[/itex] and the change in velocity is [itex]\Delta v = a (\Delta t)[/itex]. With constant velocity v it is [itex]\Delta x = v (\Delta t)[/itex].

    1) So in the first 8 seconds, it speeds up to 20 m/s. There is a formula which will give you the distance after a time t (find it in the above text). You will notice that, to apply it, you should first find the acceleration a. There is another formula which will give you that: [itex]v = a t[/itex] -- all you have to do is fill in the values that you are given (and then solve for a). Now the position graph is given by [itex]x(t) = \tfrac12 a t^2[/itex] - with the a you have just found - which is just a parabola you should be able to draw. Give it a try and post what you get.

    2) You know that the velocity is 20 m/s and it is constant. How long does it take to cross 60 meters? (I gave the formula, which one is it? Again, you have to solve for the variable you want).

    3) Now apply the formula you gave: [itex](s_1 - s_2) = \tfrac12t(v_f - v_i)[/itex]. You know at this point where the truck is (what s1 is) and you know how far it will go (it's given: s2 = 180 m). You also know at this point the velocity (it's given, what is it again?) and the final velocity (it's implied in the text). So you can calculate the time it will take to stop. Now [itex]\Delta v = a \times \Delta t[/itex] will give you the acceleration (plug in the velocity difference and time you just calculated) and you can use the same formula as in the first part, for the displacement.

    Try to solve each phase separately. The second is the easiest, the third the hardest. Just post anything you tried, no matter if you think it's nonsense. I'm trying to get you started without giving away the answer entirely, but perhaps your problem is somewhere else than I think and your attempts will maybe point that out to me.
     
    Last edited: Sep 15, 2007
  8. Sep 15, 2007 #7
    Yeah so he'd be going at a constant speed for 3secs so your graph of velocity vs time graph would have a straight horizontal linefrom t=8 until t=11. After this he starts to slow down again so you can find how long he is slowing down for by using displacement=0.5x(initial+final speeds)x time rearranged to make time the subject. You can then plot the final part of the graph (hopefully!)
     
  9. Sep 15, 2007 #8
    ok this is what i did so far,

    for phase 1

    v=at
    20= a (8)
    a= 2.5

    so the equation I will plug it from 0-8 seconds will be x=.5(2.5)(t^2)

    and for phase 2

    60=.5(2.5)(t^2) <-----I solve for t

    t = 5.42
    so it will be at x=140 at 11.8

    phase three

    180-140 = .5t(-20)
    40=10t
    t = 4

    40 = a4
    a=10???

    ....
    let me try the other approach

    110 = .5(10)(t^2)
    t = 4.69


    also

    what I drew as of now...[​IMG]
     
    Last edited: Sep 15, 2007
  10. Sep 15, 2007 #9

    learningphysics

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    Ok... step by step... first phase looks good:

    x=.5(2.5)(t^2) for 0<=t<=8

    Now the question says that from t = 8... he drives 60m at a constant speed... we know that he reaches 20m/s at the end of the first phase... so how many seconds is this phase... can you write the equation for phase 2?
     
  11. Sep 15, 2007 #10
    so he drives 60 meters.

    he is at 20m/s so that means that he will travels this distance in 3 seconds.

    x = 60m/20m/s for 8<=t<=11
     
  12. Sep 15, 2007 #11

    learningphysics

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    Here's a trick... you know that this section is constant velocity... so the formula for x should be linear:

    x = at + b

    what should a be?
     
  13. Sep 15, 2007 #12
    a is the slope.

    velocity...
     
  14. Sep 15, 2007 #13

    learningphysics

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    exactly... so for this section the equation is:

    x = 20t + b (since you know the 20m/s is the velocity here)

    what is b?
     
  15. Sep 15, 2007 #14
    usually that is the y intercept.

    since t = time and x = postion then b would be +60
     
  16. Sep 15, 2007 #15

    learningphysics

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    No... what you want to do is take a particular time, and the position at that time... plug them in and solve for b...

    x=.5(2.5)(t^2) for 0<=t<=8

    x = 20t + b... this is from 8<=t<=11.

    what you can do is find the x value at t = 8, from the first equation... plug it into the second and solve for b... both equations are valid at t=8... so they should both give the same x value at t=8.
     
  17. Sep 15, 2007 #16
    after doing so, I learned that b = -80.

    so now I continue with the equation provided until 11 seconds.

    so I got that at 9 seconds x is 100 and continued so until the 11th second.

    now for the third phase,

    the manner would be the same, but the slope will not remain the same as its decreasing. The manner of solving for b must be the same for the third phase, as again the points are shared by two equation. It will have to be another linear equation.

    140 = m(11) +b

    m has to be different (but how much different)
    and b is different.I would think 180.

    140 = 11m +180

    m=-3.63

    y=-3.63x+180?
    ??
     
    Last edited: Sep 15, 2007
  18. Sep 15, 2007 #17
    the equation I came up with before, cannot be because x is decreasing, and x cannot decrease, instead it should continue growing until 180. So much for me being happy. :(
     
  19. Sep 15, 2007 #18

    learningphysics

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    The last section is not linear (remember linear is only for constant velocity... ie 0 acceleration: x = vt + x0)... the form of the equation for the last part is:

    x = x0 + v1*t + (1/2)at^2

    we also know that v1 here is 20m/s (because the last phase was constant velocity at 20m/s, so the beginning of this phase velocity is 20)

    x = x0 + 20t + (1/2)at^2

    can you get the acceleration?
     
  20. Sep 15, 2007 #19
    180=140+20(11)+.5(a)(11^2)
    180=360+60.5a
    a=-2.97

    equation would be:

    x= 140 + 20t+.5(-2.9752)(t^2)

    ....that is if the acceleration is correct.
     
  21. Sep 15, 2007 #20

    learningphysics

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    But this is after t=11... t>11. Also, is the 180m supposed to be from the beginning of the whole thing, or from the moment he started decelerating? I'm confused by that.
     
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