How to draw a position-time graph from a velocity-time graph?

  1. Dec 28, 2010 #1
    And vice-versa? I'm having quite a hard time at doing this. Any tips? My book isn't exactly "layman" friendly. Thanks so much for all the help, I appreciate it. :smile:
     
  2. jcsd
  3. Dec 28, 2010 #2
    The trick is in derivatives!

    velocity is the derivative of position, dx/dt=v
    So all you need to do is draw the derivative graph of position-time and you will have the velocity-time graph!

    Here's a site with pretty trivial examples that might help, it gets trickier as the position-time graph gets more complicated, you have to be able to look at the functions for the graphs, take the derivative and then graph, or just estimate it.
    http://www.physics247.com/physics-homework-help/motion-graphs.php

    If you're going backward from velocity-time to position-time, you have to take the anti-derivative of the graphs, it's bascially just working backward; finding the position graph whose derivative would make the velocity-time graph you are looking at. It's a little more difficult to do that way.

    Mathematically (if you're given the functions), an example would be
    x(t) = 3t+5
    v(t) = dx(t)/dt = 3
    the value for the velocity function is simply the slope of the position function, 3, in this case, it's a linear function. For non-linear functions, it changes
    x(t) = 2t2
    v(t) = dx(t)/dt = 4t
    Now the derivative is a function.

    Hopefully that's all the info you need and more. Sorry if the derivative explanation was unneeded, didn't want to make any assumptions about your math level.
     
  4. Dec 28, 2010 #3
    Thanks for the reply, soothsayer! :biggrin:

    The problem is, I am only doing grade 11 physics at the moment and haven't even touched calculus (which is next semester) so I have no idea what derivatives are.

    But, my book basically tells me the first thing I need to do is find the displacement for every part of the graph.
    Example: say I have a velocity-time graph, and the velocity is 50km/h [forward] in 0.10 h. And now, I would need to find the displacement for every 0.01h so I could start constructing the position time-graph. How would I be able to do that?

    Thanks in advance.
     
  5. Dec 28, 2010 #4
    Ok, as long as the graphs are simple, you can do them without calculus.

    So we have a velocity-time graph where velocity is 50km/hr as you say, this velocity is sustained for 0.1 hr. So if the object in question travels at 50km/hr for 0.1 hr, the total displacement = 50 x 0.1 = 5 km, now, this only tells us the endpoints of the position-time graph, x=0 at t=0 and x=5km at t=0.1 hr, to know the rest, we have to look at the velocity-time graph bit by bit. So, pick arbitrary time values for t (a good one would be, as you mentioned, every 0.01 hrs) and figure out the displacement up to that point by multiplying those time values by the velocity.

    t = 0 hr, x = 50 x 0 = 0 km
    t = 0.01 hr, x = 50 x 0.01 = 0.5 km
    t = 0.02 hr, x = 50 x 0.02 = 1 km
    ....
    t = 0.09 hr, x = 50 x 0.09 = 4.5 km
    t = 0.1 hr, x = 50 x 0.1 = 5 km

    You'll see as you start to plot these points for position and time on a graph, that it creates a very simple linear function, so you can simply draw a straight line from the starting point (0,0) to the end point (0.1, 5) and that will be a perfect representation of the position-time function! This is the case every time velocity is constant, such as in this case, it never changes from 50km/hr

    So the method for constructing these position-time graphs from velocity is not hard, simply multiply different time values by the velocity to come up with a displacement. The displacement would be the value of the y coordinate and the time value you have chosen is the corresponding x coordinate value.
     
  6. Dec 28, 2010 #5
    That clarified many things, thanks soothsayer.:biggrin:

    Okay, now I have two graphs I'm going to have convert to position-time graphs, just want to see if I did it right:
    1. The first graph: http://img189.imageshack.us/i/kinematicsvelocitytimeg.jpg/
    The first line is 70km/h, but the time-graph is in minutes so I'll have to divide that by 60: 70/60 = 1.16 km/m. The change in time is 5 minutes for constant velocity, so the truck travels about 5.9 km south in 5 minutes at a constant velocity before it accelerates. Then the truck begins to travel at 90km/h for 30 minutes. Dividing 90km by 60 to get 1.5 km/m, and multiplying it by 30 minutes to find that the truck traveled a distance of 45km south. Total displacement so far is 50.9km south. Now the truck begins to slow down and travels at a constant speed of 80km/h which needs to be converted to km/m, which becomes 1.333 km/m. Change in time is 30 minutes again, so multiplying that by 1.333 gives us a distance of 40km traveled by the truck. The total displacement for this graph is 90.9km.

    So I made the position-time graph on paint after deducing the above:
    http://img214.imageshack.us/i/kinematics555graph.jpg/
    I made the slope of the first line (which is 70km/h) the least steep, while the second line which is 90km/h the most steep, and finally the third less steep than the second but more steep than the first. Does this look right?

    2. Now for the second graph: http://img9.imageshack.us/i/kinematicsvelocitytimeg.jpg/
    The truck travels at 70km/h for 10 minutes, since the time graph is in minutes I need to convert 70km/h into km/m and once again have a value of 1.16km/m. Therefore, the truck will travel 11.6 km in 10 minutes. Then the truck begins to pick up speed and travels at 80km/h for 30 minutes. The truck will travel a distance of 40km during this time. Total displacement so far 51.6 km. The truck then loses speed and travels at 75km/h for 5 minuets, which converts to 1.25km/m multiplied by 5 minutes, which shows that the truck traveled 6.25 km over these 5 minutes. The truck then loses more speed and travels at 60km/h, which is 1km/m in 2.5 minutes. The truck travels 2.5 km during this time. Total displacement so far is 60.35 km. The truck then deaccelerates to 50km/h for the rest of the distance which is 22.5 minutes. 0.83km/m(22.5 minutes) yields a distance of 18.75 km. The total displacement is 79.1 km.

    So, like I did with the first graph, I made a second one based on the deductions above on paint: http://img31.imageshack.us/i/kinematics666.jpg/
    I made the slopes more or less steep depending on the speed the truck traveled during those times.

    If I did anything wrong, what can I fix? Thanks for all the help thus far. :smile:
     
  7. Dec 28, 2010 #6
    Those both look perfect! Yeah, that's exactly the way to do it.
     
  8. Dec 28, 2010 #7
    Wow, thanks again soothsayer for all the help you've been. :biggrin:

    Just one more question, if I have another question that is bugging me can I post it in this thread or do I start a new one?
     
  9. Dec 28, 2010 #8
    I would start a new thread, unless the second question is related to this one, then you can post it here.
     
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