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Homework Help: Finding the x component from a velocity vs. time graph

  1. May 24, 2009 #1


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    First-time poster here.

    1. The problem statement, all variables and given/known data

    A little cat, Bella, walks in a straight line, which we shall call the x axis, with the positive direction to the right. As an observant scientist, you make measurements of her motion and construct a graph of the little feline's velocity as a function of time.

    Find the x component of Bella's velocity at t = 2.00 s. (in m/s)

    Find the x component of Bella's velocity at t = 7.00 s. (m/s)

    What is the x component of her acceleration at t = 2.00 s? (in m/s^2)

    What is the x component of her acceleration at t = 6.00 s? (m/s^2)

    What is the x component of her acceleration at t = 7.00 s? (m/s^2)

    For some reason, I can't get the image to show up. Here is the link to it: http://img32.imageshack.us/img32/4815/1012297.jpg [Broken]

    2. Relevant equations

    v = [tex]\Delta[/tex] x/ [tex]\Delta[/tex] t
    a = [tex]\Delta[/tex] v/ [tex]\Delta[/tex] t

    3. The attempt at a solution

    The graph is velocity versus time with a negative slope, and the slope is the acceleration. I have no clue as to how I should go about finding the x-component of the velocity in m/s. There's no angle given, only the graph. As for the acceleration questions, I found the the slope of the change in velocity over the time (according to the values on the graph) and the answers were also incorrect. Any hints?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 24, 2009 #2
    In what direction is she moving?
    So what should you do to get x-component?[It's the easiest thing to do in the world :P]
  4. May 24, 2009 #3
    If i'm not mistaken, your graph shows the x-component of the velocity against time - so you should be able to simply read from the values from the graph. You also must be careful as the units on the graph are cm/s - and the question is asking for m/s.

    As for the acceleration, it looks as though you've done the right thing - but once again make sure you convert from cm/s^2 to m/s^2.
  5. May 24, 2009 #4
    its difficult to aquire x components from the graph so I believe that an relevant equation would be:

    V = V(initial)+at

    where a is the slope of the graph (which, since the line is straight, is known to be constant throughout).
  6. May 26, 2009 #5


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    Hello everyone,

    I appreciate the responses. After getting the answers wrong yet again, I found a similar problem via Google. I was (sort of) on the right track by reading the values off of the graph.

    The only problem was that the answers require precision down to the decimals (i.e. the velocity for t = 2 s was 5.33 cm/s). That's obviously hard to see on such a tiny graph. The average acceleration is roughly 1.3 cm/s^2, so I would have been better off doing this calculation first, then determining the change in velocity for each point in time (just as you said, bruiser).
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