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Motion graphs and their units

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm having some issues with the following exercise:

    The graph below represents the marking of a vehicle speedometer in funtion of time. Elaborate the corresponding graphs of acceleration and space traveled in function of time. What is the average acceleration of the vehicle between [itex]t = 0[/itex] and [itex]t = 1 min[/itex]? And between [itex]t=2min[/itex] and [itex]t = 3min[/itex]?
    2. Relevant equations

    3. The attempt at a solution
    First I obtained all the equations of the straight lines, which are as follows:[tex]
    v(t)= \begin{cases} 90t, \textrm{ if } 0 \leq t \leq 0.5 \\ 45, \textrm{ if } 0.5 \leq t \leq 2 \\ -90t+225, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 150t-450, \textrm{ if } 3 \leq t \leq 3.5 \\ 75, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -150t+750, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex]I know the acceleration graph data can be obtained by derivating the velocity functions and the displacement graph by integrating them, so:[tex]
    a(t)= \begin{cases} 90, \textrm{ if } 0 \leq t \leq 0.5 \\ 0, \textrm{ if } 0.5 \leq t \leq 2 \\ -90, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 150, \textrm{ if } 3 \leq t \leq 3.5 \\ 0, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -150, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex][tex]x(t)= \begin{cases} 45t^2 + C, \textrm{ if } 0 \leq t \leq 0.5 \\ 45t + C, \textrm{ if } 0.5 \leq t \leq 2 \\ -90t^2+225t + C, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 75t^2-450t + C, \textrm{ if } 3 \leq t \leq 3.5 \\ 75t + C, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -75t^2 +750t + C, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}[/tex]And so I happily elaborated the graphs. And when I went to compare the results to the ones in my textbook, I realized there was no answer there. So I checked on the web and I found a document from an unknown source in which the question was solved, and I was faced with the following graphs:
    onnJ5yx.png J24WGEF.png
    I know they are correct in terms of how they are charted, what bothers me are the units. It doesn't make sense to me how can the acceleration be around [itex]5000m~h^{-2}[/itex], for instance. I think it has something to do with the fact that the time is being measured in minutes, and perhaps I'm not converting it properly (I transformed minutes to seconds when finding the acceleration and position, and then converted it back to minutes. Am I being clear enough? D:) but I'm not so sure about that.

    So, how do I convert velocity [itex](\frac{km}{h})[/itex] to position and acceleration? Any help to understand this would be greatly appreciated!
    Last edited: Apr 21, 2017
  2. jcsd
  3. Apr 21, 2017 #2
    Isn't average acceleration just the change in velocity divided by the change in time? So what is the change in velocity in the first minute? And what is the change in time in the first minute? I get (45 km/hr)/(1 min) = 45 km h-1 min-1
    Now I can convert that to whatever units I choose. km/hr^2 or m/s^2, or nm/decade^2. All you have to do is know how to convert units, true?
  4. Apr 21, 2017 #3
    Yes, that's true. So [itex]45km~h^{-1}min^{-1} = 2700km~h^{-2}[/itex]. Why does it say around [itex]5000km~h^{-2}[/itex] in that graph? Is it incorrect?
  5. Apr 21, 2017 #4
    Basically, the graph is showing instantaneous acceleration. So if you average it over the first 30 seconds, you will get 5400 km/hr^2. But if you average it over the first 1 minute, you will get half that value, or 2700 km/hr^2, because the latter 30 seconds has 0 acceleration. Does that make sense?
  6. Apr 21, 2017 #5


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    If it was me doing this problem, I would convert everything to a common unit system. The choice is yours - as long as everything is consistent, then you should arrive at the correct answers. For example, suppose we want to do meters per sec for velocity, and seconds, for time. For the first one, you need to convert 45 km/h to meters/sec. If you travel 1 kilometer in 1 hour (1 km/h), then you have traveled 1000 meters in 3600 sec, so if you take the stated v (in km/h) and divide by 3.6 you will have m/s. So 45 km/h becomes 12.5 m/s, So the first line is v(t) = (0.416667 m/s^2)*t, for 0s < t < 30s. and go on like that.

    Now if we convert everything to km and hours, you have v(t) = 5400 km/h^2 from t = 0h to (1/120) h, for the first line, which matches the first part of the graph.

    One thing to remember about acceleration: Think of 1 m/s^2 as (1 meters per second) per second. So each second, it increases by 1 meters/sec, then the next second it gains another 1 m/s, etc. So if in 30 seconds, you go from 0 to 45 km/h, by 1 minute (at same accel) you will be at 90 km/h. Keep going at this same rate of acceleration, and you will be at 5400 km/h after 1 hour. This video (just in the first 2 minutes of it) explains about acceleration nicely, but the whole video is worth watching, when you have time.
    Last edited: Apr 21, 2017
  7. Apr 21, 2017 #6
    Wow. That makes it very clear for me now. Thank you!
    I have another question, though. What is the value that [itex]C[/itex] must assume (from the integration) when I'm sketching the position graph?
  8. Apr 21, 2017 #7
    I guess to figure that out, you have to know the x position at t=0, because at t = 0, x=C. So what is the position at t=0? If x=0 at t=0, then C has to equal 0.
  9. Apr 21, 2017 #8
    Okay. Got it. Thank you very much! :3
  10. Apr 21, 2017 #9
    No. Actually I didn't get it, haha. Take for instance [itex]t = 3[/itex]. It'll result in a negative position (against the trajectory), according to the equation of the straight line, but that's not what it looks like in the reference graph. Am I missing something...?
    Last edited: Apr 21, 2017
  11. Apr 22, 2017 #10
    Sorry it took me so long to get back with you. I was away. So in the original plot, from t = 2.5 to t = 3, you have v(t) = 0. Given v(t), to find x(t) you have to integrate. What is the integral of 0? It is C, not zero.

    Think about a car moving according to the original velocity graph. It drives in a straight line. It starts from a stop and accelerates up to 45 km/h and stays at 45 km/h for a minute and a half. Then it slows down until it reaches 0 km/h. It has moved quite a distance from its starting point, true? It didn't somehow jump back to x=0 when it stopped. After it is stopped for 30 seconds it is still at that same position. Then it accelerates again, moving even further from its starting point, and eventually stops further away still. I hope that makes sense.
  12. Apr 22, 2017 #11
    Oh. I can understand that. But when I said [itex]t = 3[/itex], I was talking about this line
    It's just that you said that [itex]C = 0[/itex], but that's just in the first segment, right? So in the next segments, [itex]C[/itex] is determined by the previous segment's final velocity?

    By the way, a correction since I can't edit my original post, in the third function it should be [itex]45t^2[/itex].
  13. Apr 22, 2017 #12
    I'm sorry; I was kind of rushed. You are right. C must be calculated individually for each segment. So for segment 1, it makes sense to define x=0 for t=0. Then, if I did my math right, it looks like at the end of segment 1, the x position is 11.25. So for the next segment (segment 2), you can find C for that equation by plugging in t=0.5 (the starting time of the second segment) and the known position of 11.25 (the starting position of the second segment).

    One other thing I was thinking about was this: I know that if I accelerate hard in my car, my speed increases at a rate of about 10 mph in 1 second. So although people don't normally talk in these terms, the acceleration is 10 mph/second.
    Now what would the speed change be in a 10 second period? Well it would be 10 times 10 mph, or 100 mph.
    Or you could say that the acceleration is 100 mph/10seconds.
    And if my car was able to, what would the acceleration be in a minute. Well, it would be (60) x (10 mph/second), or 600 mph/minute, or 3600 mph/hour, written, as you know, as 3600 mi/hr^2.

    One other note: The problem with my car is that, for various reasons, it is not able to maintain that 10 mph/second acceleration for very long. :(

    Last note: Good catch on your correction.
  14. Apr 22, 2017 #13
    Yes, that does make sense, I think I understand the acceleration part already, but I can't translate it in terms of the position... ):

    Anyway, I still don't get what the functions should be. For the second segment I tried [itex]t = 2[/itex], so it ended up as [itex]45 \times 2 + 11.25[/itex], but that doesn't look like right, it's highly inconsistent with the answer graph. So my guess is that perhaps what you want to say is that I must consider each segment as if it were the start of the graph in terms of the function, but adding the [itex]C[/itex] to represent the continuity of the graph? Am I being clear enough? D:
    Last edited: Apr 22, 2017
  15. Apr 22, 2017 #14
    You wrote: 45x2 + 11.25 for t = 2. That is not right. It should be 45x2 - 11.25.

    For the first segment, you found the equation: x = 45t2 + C, if 0≤t≤0.5
    That looks right. So, at t=0, you know that x=0. So setting 45(0)2 + C = 0 results in C = 0 for segment 1, right?

    For the second segment, you integrated the velocity equation to find: x = 45t + C
    And you know that at t = 0.5 seconds, the position is 11.25 m, because 11.25 m was the final position for segment 1. So that has to be the position at the start of segment 2.
    So plugging in the values for x and t at the start of segment 2, you have: 11.25 = 45(0.5) + C
    Solving that results in C = -11.25 for segment 2. So your final equation for segment 2 should be x = 45t - 11.25 (that is, if I did my math right)

    And continue in that manner for the reset of the segments. Make sense?
  16. Apr 22, 2017 #15
    Oh! That makes a lot of sense to me now!
    So, for [itex]t=2[/itex], I have [itex]90 - 11.25[/itex] which is [itex]78.75[/itex]. But that is obviously not in minutes, so I'm gonna guess I have to divide it by [itex]60[/itex]...?
  17. Apr 22, 2017 #16
    No, do not divide by 60. The value 78.75 is not time; it is the value for x. You found that at t = 2, the position of the vehicle is 78.75 meters from the start.
    So now that you know the position at the end of segment 2, that is also the position at the start of segment 3. Using that value for x, and using the time at the start of segment 3 to be t = 2, you can plug those values into the position equation for segment 3 and solve it to find the value of C for segment 3.
  18. Apr 22, 2017 #17
    No, I mean, I know that what I found is not time, but [itex]78.75[/itex] is not the properly represented position (in [itex]km[/itex]) in the answer graph at [itex]t = 2[/itex] (it is something between [itex]1km[/itex] and [itex]2km[/itex]), so I'm guessing there is some sort of convertion I am missing, which applies to all of the other segments aswell.
  19. Apr 22, 2017 #18
    OH, I am so sorry. I mixed units. I was mixing hours and minutes because that is how the chart is drawn: km/hr on the y axis and minutes on the x axis. My math is completely wrong on this problem.
    Let me shout it from the rooftops. YOU ARE CORRECT!!! I AM WRONG!!!
    I am really sorry. Yes, you have to be on the same units.

    So I think that somehow you should probably change all of your equations so that units are matching.

    So for segment 1, for example, we found the equation to be: x = 45t2 + C
    If you want to keep time in minutes, then because 45 km/h = 0.75 km/minute, the equation should be:
    x = 0.75t2 + C (At least that's what I think it should be.)
  20. Apr 22, 2017 #19
    Hahahahaha. It's OK! That's very confusing anyway!

    So, from the first function, we have now that for the second segment [itex]C = 0.1875[/itex]. Which means that for [itex]t = 2[/itex], the position will be [itex]0.75 \times 2 - 0.1875 = 1,3125[/itex], which seems to match the answer graph. Is that right?
  21. Apr 22, 2017 #20
    That looks right, based on a very quick calculation. I'm sorry I don't have any more time to spend right now. I would really like to in order to bail myself out. But I have to run my neighbor to the airport right now. Once again, I'm sorry for the bad information I gave you. But I got the same answer of 1.3125 at t = 2.
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