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Reading an Acceleration Vs. Time Graph

  1. May 1, 2008 #1
    1. I need to find the velocity of an object with a constant acceleration of 0.5 ft/sec by creating an acceleration vs. time graph spanning the interval of 0 to 5 seconds

    2. This is kind of a theoretical problem, so if any equations apply, they were not given to us

    3. I have plotted acceleration vs. time, which in this instance is a straight line, and velocity vs. time, which is linear with a constant slope. My question is this: if acceleration is constant, and velocity is the area under that "curve" ( this is a little confusing, as the assignment calls this a curve and it doesn't look like a curve to me!), can I multiply the X coordinate by Y, which =0.5, to get my velocity at various times along this interval? This seems to make sense, as the areas under the "curve" are rectangles given the constant acceleration, but it seems too easy to be right!
    Last edited: May 1, 2008
  2. jcsd
  3. May 1, 2008 #2
    When physicists/mathematicians use the term "curve" they often mean any plot on a graph / any continuous line (straight, or curved).
    Velocity is the area underneath the acceleration "curve." Similarly, distance is the area underneath the velocity "curve."
    To find the velocity at a point in time, you find the area of the given rectangle (from time = 0 until whatever time you are loking for). To find the displacement (position) at a point in time, you find the area of the triangle in the velocity graph.

    The process of finding the area under a graph leads to what is called an Integral --> its the precise, mathematical way of finding the area under any shape. Using an integral, you can more easily convert from acceleration to velocity; velocity to distance; or in general convert from the change of something -> to what that thing actual is.
  4. May 1, 2008 #3
    Yes, you are correct in your method. This is a good intro to basic integral calculus. See if you can follow this:

    In your case your acceleration is constant, so the area under the curve is the area of a rectangle.

    But what if your acceleration was a(t)=t^2 and you want to know the change in velocity on the time interval [0,5]? Then the acceleration is not a uniform line, so we cannot just find the area using a rectangle.

    So how are you going to compute the area under that curve? Well you can still approximate the area under that acceleration curve by making a rectangle, however it's not a very good approximation.

    What if instead to split the portions of the graph up into 5 smaller rectangles, each with a width of "1 second". Now if we add these smaller rectangles to get the collective area we will find that they approximate the acceleration a little better.

    Now say you split up the rectangles up into 10 rectangles and add them instead of 5, now your approximation is even better. What about adding 50 small rectangles? Now it's a lot better. This is the basic principal of integral calculus.

    Here's the leap: What if we shrink the width of those rectangles to an really, really tiny width? Say we make the width of the rectangles infinitely small in their width, now you have an approximation that is so good it's not an approximation anymore, but the actual, exact area under that acceleration curve.

    Analyzing graphs that do not have uniform areas becomes somewhat trivial with the help of calculus, this is one reason why it is so valuable.

    Using some notation it would look like this:

    [tex]\int_{0}^{5}t^2 dt[/tex] Your area is length * width so your length is x^2, and your width is an infinitely small dx. Basically it's the process of summing up a infinite number of infinitely small pieces of area.
    Last edited: May 1, 2008
  5. May 1, 2008 #4

    Thanks for helping! I just started doing integrals and Reimann sums a couple of weeks ago in calculus, but the problem I'm working on is for an engineering class. I thought that since we are approximating rectangles to find approximations of the area under the curve in calculus, if the area was rectangular in shape already that I could skip the calculus and just take the area of the rectangle itself. Happily for me, that appears to be the case!

    Thanks again!
  6. May 1, 2008 #5
    Yeah if you can find the area an easier way, normally in the shape of a trapezoid or a rectangle, then you're free to do that, it's the same thing as the integral, just easier.

    You took the area 0.5*5

    The integral is just

    [tex]\int_{0}^{5}0.5 dt = 0.5*5[/tex]
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