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Math Help - Kinematics graph help!

  1. #1
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    Kinematics graph help!

    Generally, these are easy to figure out... but this one is really awkward.

    As you can see, the speed is in km/h and the time is 25 min, and the graph itsself is very awkward :/
    If anyone could explain to me how to get the distance, I would appreciate it.
    The answers are 32.5 km and 78 km/h.
    I had them down in class but I have no idea how my teacher did it and now the final examinations are only a day away.
    Any help would be appreciated.
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  2. #2
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    Hi

    I am afraid the answers are not correct
    The usual way to manage this problem is to come back to the definition of the speed
    v(t) = \frac{dx}{dt}

    dx = v(t) dt

    And integrate on the 3 parts (for instance on the first part, v(t) is linear v(t) = at+b)

    In your exercise you can do a little bit quicker due to the fact that the speed is linear with respect to time
    On the first part it is equivalent to the average speed (70 km/h) during 10 minutes
    On the third part it is equivalent to the average speed (70 km/h) during 5 minutes
    Therefore the global is equivalent to
    70 km/h during 15 minutes = 17.5km
    20 km/h during 10 minutes = 3.33 km
    The total distance is 20.83 km

    The answers are right only if there is a typing error on the lowest speed : if you replace 20 km/h by 60 km/h then it is true
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  3. #3
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    Quote Originally Posted by xcalibur View Post
    Generally, these are easy to figure out... but this one is really awkward.

    As you can see, the speed is in km/h and the time is 25 min, and the graph itsself is very awkward :/
    If anyone could explain to me how to get the distance, I would appreciate it.
    The answers are 32.5 km and 78 km/h.
    I had them down in class but I have no idea how my teacher did it and now the final examinations are only a day away.
    Any help would be appreciated.
    You should know that the area under a velocity-time graph gives you the displacement.

    In your graph it appears that the object is always travelling in the same direction so the graph is essentially a velocity-time graph. And since the velocity is always positive the displacement will equal the distance travelled. So .....

    The first thing I suggest that you do is convert the speeds into the unit of km/min. Now calculate the area under the graph to get the displacement and hence distance travelled (in units of km).


    Edit: Beaten by running-gag. But that's OK ..... We've proposed different methods. NB: We get the same answer.
    Last edited by mr fantastic; December 13th 2008 at 02:23 AM.
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  4. #4
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    Quote Originally Posted by running-gag View Post
    The answers are right only if there is a typing error on the lowest speed : if you replace 20 km/h by 60 km/h then it is true
    Yeah, my bad I messed up the diagram.. the speed is 60. I knew I should have just taken a pic with my cell phone.
    The problem I am having with is calculating the area in the graph ( this is the way I'm supposed to be doing it anyway ) this is How I'm attempting it and the answer is incorrect obviously.

    Would appreciate if someone could let me know how to calculate the area under this specific type of graph.
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  5. #5
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    Quote Originally Posted by xcalibur View Post
    Yeah, my bad I messed up the diagram.. the speed is 60. I knew I should have just taken a pic with my cell phone.
    The problem I am having with is calculating the area in the graph ( this is the way I'm supposed to be doing it anyway ) this is How I'm attempting it and the answer is incorrect obviously.

    Would appreciate if someone could let me know how to calculate the area under this specific type of graph.
    There are three areas under the curve. Two are trapeziums and one is a rectangle.
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  6. #6
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    Got it, thanks for the help guys.
    Totally forgot its supposed to be under the curve.
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