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Help with translating a D-T graph into a V-T graph

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi. For this assignment, I was told to create a V-T graph with my own values (this doesn't have to make sense in the real world). The assignment says that a ball rolls on the floor, up a ramp, back down the ramp, and backwards along the floor. Here is the diagram: https://tcdsb.elearningontario.ca/content/enforced2/4140033-EL_SCI_SPH3U1-21_981230_Summer1314/SPH3UCU01/SPH3UCU01A03/images/image12.jpeg?_&d2lSessionVal=4vs9HNh5MsF4f7V1pZib3bJfw&ou=4140033

    Here is my diagram with values (again it doesn't have to make real world sense):

    UxQNnEJ.png

    Here is my D-T graph:

    96iZ6y2.png

    Now I must turn it into a V-T graph.


    2. The attempt at a solution

    Here is my failed attempt at making it into a V-T graph:

    FxmPumh.png

    I know it is wrong. I know from 0-8 seconds, that portion of the graph is right. The velocity from 0 to 8 seconds is constant. The ball is rolling at 3 m/s. But after that I start to get confused. From 8 to 14 seconds, the velocity is never constant (I don't think). It is curved at every instance. The same goes from 14 seconds to 17 seconds. It is also curved like the portion from 8-14 seconds, but this time it is moving backwards, and at a faster velocity. I think the velocity from 17-19 seconds, however, is constant. The rise/run is exactly -12, so I think that the line from 17-19 seconds should instead be flat.

    I think I've covered everything I vaguely know, but I still don't know how to turn this thing into a proper V-T graph. If anyone could help, it'd be greatly appreciated, thanks.
     
  2. jcsd
  3. Jul 8, 2014 #2

    HallsofIvy

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    Assuming no friction, the speed on the level part will be constant. If it goes 24 meters in 8 seconds, that speed is 24/8 = 3 m/s. As it goes up the slope, there will be a constant "de"celeration (negative acceleration) so the distance is a quadratic function of time. One way to get that is to write "a" as the constant acceleration. The vertical speed at any time is at+ 3 m/s and the vertical distance is [tex]at^2/2+ 3t[/tex]. To find a, use the fact that from t= 8 to t= 14, the ball has traveled 16 meters.

    On the trip back, the acceleration is -a. Use that to determine the speed at the bottom of the slope.
     
  4. Jul 8, 2014 #3

    Orodruin

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    I suggest first checking whether or not the ball actually reaches the top of the ramp... The big question in my mind is how to interpret "it does not have to make real world sense". Either it follows Newton's laws or I can draw an arbitrary path as long as it passes through the given points...
     
  5. Jul 9, 2014 #4

    ehild

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    According to the picture, the ball reaches the top in 8 s. The initial speed was 3 m/s, it stops after travelling 16 m in 6 s. You can assume constant (negative) acceleration. When travelling backwards it reaches the bottom in
    3 s and then travels 24 m in 2 s, which means 12 m/s speed, so it gained energy from nothing. It has no sense in the real word.

    ehild
     
  6. Jul 9, 2014 #5

    collinsmark

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    Hello in31l,

    Welcome to Physics Forums! :smile:

    As others have pointed out, this is a strange problem that either involves other forces besides gravity, or it could be that the ramp itself (thus the frame of reference) is moving.

    [STRIKE]Also, the "velocity vs. time" graph should probably be retitled as "speed vs. time."[/STRIKE]

    Anyway, to find the corresponding points on the [STRIKE]speed[/STRIKE] velocity vs. time graph, you need to find the slope of the line on the position vs. time graph.

    I see problems with your chosen [STRIKE]speed[/STRIKE] velocity vs. time points at t = 14 s, and t = 17 s. What are the slopes of the position vs. time line at these points?
     
    Last edited: Jul 9, 2014
  7. Jul 9, 2014 #6
    Thanks. This is really the first question I've ever done involving physics (it's grade 11 physics, by the way), so I'm unfamiliar with some of the physics/math jargon being used in this thread. And about it not making sense in the real world, I mean that in every sense. As in, these are just numbers, no friction, real gravity, anything. Purely these numbers. Our teacher told us to use any values we want, it doesn't necessarily have to make any sense.

    I made a graph using simpler points, does this look correct to you?

    GEhXeeJ.png

    Also, why should I rename it to a speed vs time graph? Isn't velocity on the y-axis, and time on the x-axis, therefore a velocity/time graph?
     
  8. Jul 9, 2014 #7
    You're correct. It makes no sense, and in this case, it doesn't have to.
     
  9. Jul 9, 2014 #8

    collinsmark

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    Okay. That explains it. :smile:

    That looks more reasonable to me. :approve: Realize though that we're making a few assumptions such as the ball actually reaches its peak position at t = 7.5 s, and not some other point that isn't shown on the graph. (And it might be difficult to know the speed at the end-points of the curve.) But if you can argue your logic correctly, that looks reasonable.

    [STRIKE]I just say that because speed is the magnitude of the velocity. Velocity is a vector and has both a magnitude (speed) and direction. In this case it has both horizontal and vertical components. This is similar to the difference between displacement (a vector) and distance (a scalar).

    But if I'm guessing correctly, your teacher wants you to do this problem in terms of distance and speed (all scalars).

    (By the way, a "scalar" doesn't have a direction. It's just a number. [Although it can be positive or negative. I'm just saying that it doesn't have both horizontal and vertical components.])[/STRIKE]

    [Edit: See ehild's post below. What I said above (crossed out) is misleading. I apologize for the confusion. :redface:]
     
    Last edited: Jul 9, 2014
  10. Jul 9, 2014 #9
    Is there a way I can make it so that it is shown as a velocity? Should I add [N] on the y-axis title? I understand the difference between speed & velocity, but isn't a V-T graph (like one that shows acceleration) only a V-T graph if the y-axis is velocity?
     
  11. Jul 9, 2014 #10

    collinsmark

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    [STRIKE]Calling it velocity is fine here as long as it is understood that what is being talked about is the magnitude. Maybe I was being overly nit-picky.[/STRIKE]

    [Edit: I take back what I said about calling it "speed." Velocity is better after all. See ehild's post below.]
     
    Last edited: Jul 9, 2014
  12. Jul 9, 2014 #11

    ehild

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    You can consider the motion one-dimensional, along the route, like a train moving on the rails. It can travel forward and backward, and then you can consider the velocity positive when it moves forward and negative when it moves back.

    Speed is magnitude of the velocity so it is always positive. The speed is the distance travelled in unit time, while velocity is the displacement (a vector) in unit time. The "distance travelled" always increases.

    When you plot position in terms of time, the position is also along the route. You only need to define a positive direction. The ball moves on the horizontal section, then uphill, in positive direction, then it turns and moves backward. The final position of the ball is the same as the initial one. When you travel along a road the milestones show your position. It does not matter if the road is horizontal or leads up or down on a slope, the milestones show how far you reached along the road. So you can write "position" on the y axis in miles or km-s or meters. When plotting the velocity-time graph you can write velocity on the y axis. It is positive when the ball moves forward and negative when moving backwards.
     
  13. Jul 9, 2014 #12

    collinsmark

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    Right. Ehild said it better than I did. :redface:
     
  14. Jul 9, 2014 #13
    Thank you.
     
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