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Finding the x component from a velocity vs. time graph

  • Thread starter ran
  • Start date
  • #1
ran
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First-time poster here.

Homework Statement



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]

Homework Equations



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

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?
 
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Answers and Replies

  • #2
26
0
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]
 
  • #3
161
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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.
 
  • #4
27
0
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).
 
  • #5
ran
3
0
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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