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Homework Help: Questions on kinematics (distance/velocity/acceleration)

  1. Sep 13, 2005 #1


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    these should be pretty simple but for some reason i cant get them.

    ok they give you a basic time vs distance table:

    • 0 sec - 0 m
    • 1 sec - 2 m
    • 2 sec - 8 m
    • 3 sec - 18 m
    • 4 sec - 32 m
    • 5 sec - 50 m

    its a ball rolling down a hill which makes sense for the data. now the question is what is the distance after 2.2 seconds. can you figure that out exactly or do you just look at a graph and estimate it?

    second question is what is the slope at 3 seconds? now i konw slope is change in y over change in x (meaning change in distance over change in time) which is also the velocity at 3 seconds but which two points do you use? do you use the 3 second data with the 2 second data or the 3 second data with the 4 second data? they give different answers.

    third, if you do look at all the differences from one second to the next, you get a patten of how much it jumps - 2, 6, 10, 14, 18 - those are the velocities at each second, right? well since it jumps by 4 each time can i then say the acceleration is 4 m/s^2?

  2. jcsd
  3. Sep 13, 2005 #2
    s = .5at2 + v0t + s0

    Plug in!
  4. Sep 13, 2005 #3


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    ok so i can do this:

    s = .5(4)(2.2)^2 + (0)(2.2) + 0

    is that correct?

    am i right the a=4?
  5. Sep 13, 2005 #4
    Yes, because 2=.5a -> a=4 m/s2.
  6. Sep 14, 2005 #5


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    anyone know this question?
  7. Sep 15, 2005 #6


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    Homework Helper

    [tex]v_{f} = v_{i} + at [/tex]
  8. Sep 15, 2005 #7


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    i dont think that answers my question.

    the question specifically asked for slope (which in turn, is velocity) at 3 seconds. to find slope you need two points, right? well do you use 2 and 3 seconds OR 3 and 4 seconds to find the slope at that point?
  9. Sep 15, 2005 #8


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    Science Advisor

    You are asked, apparently, for an "instantaneous" speed and you don't have enough information to do that. To find instantaneous speed (unless the distance-time graph is a line and speed is constant), you would have to know the distance-time function and you only have it at specific times. What you can do is anyone of what you are suggesting- use the distances at 3 and 4 seconds to get the "average speed between t=3 and t= 4", (32-18)/1= 14 m/s, use the distances at 2 and 3 seconds to get the "average speed between t= 2 and t= 3,(18-8)/1= 10 m/s, or even average those two answers- (10+ 14)/2= 12. For this particular problem, that turns out to be the same as if you find the "average speed between t= 2 and t= 4", symmetric about t= 3, (38- 8)/2= 24/2= 12 m/s. If you want to do a lot of work to get a very accurate answer, you might find a polynomial function that fits all of the points and differentiate that! Actually, here, it is not a "lot of work". It look pretty obvious that each distance, in m, is equal to twice the time, in s, squared: d= 2t2.
    The instantaneous speed, at t= 3, is the derivative of that, 4t, evaluated at t= 3: 4(3)= 12 m/s.
    Of course, that was to be expected. Something moving under gravity has a constant acceleration and so its distance function is a quadratic. That was what apmcavoy was telling you in the first reply to your post.
  10. Sep 15, 2005 #9


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    ah. thanks!
  11. Sep 15, 2005 #10
    Solving the slope

    I don't think that you need to take the derivative especially if you are not in a calculus based physics course.
    To find the slope at one point draw a tangent line. I promise there is an explantion of how to do this in your book. It should just touch the curve. The tangent line is straight so take any 2 points and find the slope.
    Last edited: Sep 16, 2005
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