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Math Help - Finding the distance from a speed-time graph

  1. #1
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    (Solved) Finding the distance from a speed-time graph

    It's kinda more complicated than you might think from the title...
    the question asks:

    Calculate the distance travelled by the car while it was travelling faster than the truck.
    (the graph is attached)
    and here is my attempt:
    area of left triange= 0.5 * 30 * 24 = 360m
    area of right triange= 0.5 * 10 * 24 = 120m
    so that area where the speed in highter is 360+120= 480m
    but the answer should be 960m.

    thanks in advance :-)
    btw this forum ROCKS!
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    Last edited by IBstudent; May 8th 2011 at 02:05 AM.
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  2. #2
    Newbie Auri's Avatar
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    Wait, I'm sorry let me start from the beginning and rework this problem out.

    Okay, I understand it now. The car is traveling faster than the truck between x equals 15, and 55. Which makes your domain or your base 40. The height reaches it's max when x equals 45. That's where your triangles are split in half. Between 45, and 55, your base is 10 and the height is 24. 10*24 is 240. Between 15, and 45 your base is 30, and the height is 24 again. 30*24 equals to 720. 720+240 is 960.

    Oh wow I just figured it out!

    The whole base of the triangle is 60, the max height is 36. 36*30 equals 1080. But we only want the car, we know the truck is traveling faster between 0-15, and 55-60. That makes two right triangles. We take the area of both triangles. 7.5*12 equals 90. That's the area of the first triangle. 2.5*12 equals 30. That's the area of the second triangle. The area of both Triangles of the truck becomes 120. The whole area was 1080. Take the area of the truck from the area of the whole triangle which leaves you with the area of just the car. 1080-120 equals to 960.
    Last edited by mr fantastic; May 7th 2011 at 11:27 PM. Reason: Merged posts.
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  3. #3
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    You did it!
    so in over words...... you find the area not only above the speed of the truck, but also under it......
    thanks
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  4. #4
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    Quote Originally Posted by IBstudent View Post
    It's kinda more complicated than you might think from the title...
    the question asks:

    Calculate the distance travelled by the car while it was travelling faster than the truck.
    (the graph is attached)
    and here is my attempt:
    area of left triange= 0.5 * 30 * 24 = 360m
    area of right triange= 0.5 * 10 * 24 = 120m
    so that area where the speed in highter is 360+120= 480m
    but the answer should be 960m.

    thanks in advance :-)
    btw this forum ROCKS!
    Новый точечный &#1.zip
    Here is an attached answer.
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  5. #5
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    Quote Originally Posted by ihavenonick View Post
    Новый точечный &#1.zip
    Here is an attached answer.
    thanks
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