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Homework Help: Investigation-Help required

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data
    An item is dropped from a building and its equation is given by s(t)=5t^2
    0<(or equal to)t<(or equal to)5.

    1.)Draw a graph of this.

    2.)Find the height of the building.

    The person performing the experiment wishes to calculate the speed at which it hits the ground. He decides he will try Velocity=distance/time.

    3.) Calculate the average speed on the item's journey, strictly using v=s/t (even though it is wrong)

    He identifies that his formula ism't correct and so attempts to find the speed at each time interval of t=0 and t=1, then t=1 and t=2, and so on up to the between t=4 and t=5

    4.) What is the found speed for the intervals.

    2. Relevant equations
    I think most of this would be done on Graph mode, on a graphics calculator.

    3. The attempt at a solution

    I got 125m for the height, 25ms-1 for the overall journey, although i can't remember my answers for the next questions. For the time intervals, i entered Y-cal for the height and entered the distance between each time interval and divided it by the second in between-I'm not sure about this.

    Thankyou for your help. This is very easy, but it's hard to do something that leads you through doing the task incorrectly. Basically in these investigations you have to follow the instructions, and in this case you have to follow the student doing the wrong formulas, etc. Although as the paper unfolds the student finds the correct formula.
  2. jcsd
  3. Mar 17, 2008 #2
    Well, for 4), he measures the distance covered in each second. Say, then, between the second and first second. The distanced travelled will be s(2) - s(1), and similarly for the other ones.
  4. Mar 18, 2008 #3


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    Science Advisor

    It would be very helpful if you were to write the problem just as it is given. "An item is dropped from a building and its equation is given by s(t)=5t^2 0<(or equal to)t<(or equal to)5." doesn't make sense. "It" refers to the "item" and an "item" doesn't have an equation! Presumably this is supposed to be "the equation of its height is" except that also doesn't make sense- the "dropped item" would be going up! I have to conclude that the given equation is for the distance (In what units? You use meters later- it would have been nice to tell us that initially) below the top of the building after time t (apparently in seconds. Again you did not say that as part of the problem). Finally, there is nothing said about the "0[itex]\le[/itex] t[itex]\le[/itex]5" meaning that the item hit the ground in 5 (seconds).

    Assuming that your distance units are meters and time units are seconds, yes 125 m for the height and 24 m/s for the average velocity are correct. I have no idea what you mean by "Y-cal"- that's not a standard term and there is no mention of it in the problem.

    It's not clear to me what you are asking! Yes, what you at the end looks correct. You can find f(0), f(1), f(2), f(3), f(4), and f(5). The differences, f(1)- f(0), f(2)- f(1), f(3)- f(2), f(4)- f(3), and f(5)- f(4) give the distance the item fell during that time. Dividing by "1", the time difference gives the average speed during that second.
  5. Mar 18, 2008 #4
    That doesn't make much sense.

    Avg. Speed = [tex]|v| = \frac{s_f-s_0}{t_f-t_0}[/tex]

    Which is what you stated, essentially. Don't know why your teacher would say that's wrong for an average speed.

    Now using the time intervals might give you a better approximation to the instantaneous speed between the intervals....
    Last edited: Mar 18, 2008
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