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Homework Help: Units of the integral (x=seconds, y=metres)

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm plotting a man jumping over a car for maths and i'm trying to find the critical velocity at which the car has to travel for the jumper to make it safeley over the car the man's jump is approximated by the parabola y= -0.0176x^2 + 0.3595x - 0.2794 and restricted by the height of the car y=1.225

    So basically I have units of metres on the y-axis and seconds on the x-axis and i want to find the area under the curve of y= -0.0176x^2 + 0.3595x - 0.2794 and restricted by the line y=1.225. However i know how to do all the Integral stuff the only trouble I'm having is with what the units will be for my answer

    2. Relevant equations
    units of y axis = metres
    Units of x axis = seconds

    3. The attempt at a solution
    My best guess is the units will be m/s which is what will be of most use to me, however if it's not is there a way i can manipulate the graph to give me an answer in metre's per second i.e instead of t make it t^2 or somthing like that
  2. jcsd
  3. Mar 16, 2012 #2
    In a graph of meters vs seconds, the slope at any point (the derivative, in this case velocity) will be expressed as simply meters/second [m/s]
  4. Mar 16, 2012 #3
    Yes but that will give me the instantaneous velocity of the jumper won't it? I think I'm trying to be too tricky with this problem
  5. Mar 16, 2012 #4
    Well, velocity is always m/s, whether it's critical, instantaneous, or the magnitude (ie. speed).

    But I think I understand what you are getting at. Suppose you had a graph with m on the y axis and s on the x axis. You want to know what the unit would be for the area under the graph?

    I guess you can just look at the graph from a geometric standpoint. If the integral is essentially adding up the area under the graph, let's consider a graph like this:


    Only let's say that the y axis is in some unit m and the x axis is in some unit s. Geometrically, when you add up the area under that graph using the formula for area of a rectangle (L*W), what will the units be?

    Now recall, that during integration, you are simply separating the space under a curve into little segments dx (well, ds in this case) and multiplying that by the height (y, or m in this case). Can you see what the units would be now?
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