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IAmCreature

  • 4 years ago

Can someone please help me with this? I dont need an answer i just need some explanation on how to do it i guess :) 1. Describe what the graph shows. https://www.connexus.com/content/media/93289-4142005-104123-AM-1766211605.gif

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  1. Callisto
    • 4 years ago
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    As the absolute value of slope increases, the velocity (or speed) increase. Should I do it like this?

  2. experimentX
    • 4 years ago
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    find the slope, that will be your velocity. you have two different slopes here. so two different velocities. from 0 to 8 be velocity 1 and 8 to 10 be velocity 2. calculate them.

  3. UofIMechEng
    • 4 years ago
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    FInd the integral or the area under the curve and you will have total distance.

  4. UofIMechEng
    • 4 years ago
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    Callisto speed and velocity are different don't confuse them. The absolute value is speed and without absolute value you get velocity.

  5. UofIMechEng
    • 4 years ago
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    Velocity is a vector speed is a magnitude.

  6. Callisto
    • 4 years ago
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    To my knowledge, area under this graph is NOT the total distance travelled. Or it's better for me to amend my answer . For t=0s - 8s velocity = (6-9)/(8-0) = -0.375 For t=8 -11s velocity = (0-6)/(11-8) = -2 Take the absolute value You'll see the the rate of travelling increases.

  7. UofIMechEng
    • 4 years ago
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    Area under the graph is equal to the Integral of velocity, which is displacement/ position. Which will give you the total are equalling total distance.

  8. Callisto
    • 4 years ago
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    Please see the axis, it's y=distance , x=time, it's NOT y=velocity , x=time!

  9. UofIMechEng
    • 4 years ago
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    in this case velocity is Dx/dt = velocity. In the x direction.

  10. UofIMechEng
    • 4 years ago
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    It's the same it's only giving you the distance throughout the total time.

  11. UofIMechEng
    • 4 years ago
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    Stop and think. If the line is the rate at which the person is moving. Then the area under the rate is distance.

  12. Callisto
    • 4 years ago
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    Perhaps you should read this http://answers.yahoo.com/question/index?qid=20080906075747AABsnoO

  13. UofIMechEng
    • 4 years ago
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    Yes view said article answer under the chosen answer, that guy is correct. It is total distance traveled since there is no negative distance.

  14. UofIMechEng
    • 4 years ago
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    Displacement is when you have both negative and positive distances traveled. Please hop off my case bro. I'm too pro.

  15. amistre64
    • 4 years ago
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    @UofIMechEng , dont get a bee in your bonnet :) Noone was "on your case". Callisto was just trying to clarify their postion to avoid confusion. And yes, area under the velocity curve is displacement - given that we measure it in a straight line right?

  16. UofIMechEng
    • 4 years ago
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    Displacement as a straight line yes.

  17. amistre64
    • 4 years ago
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    But to answer the question; im not really sure what the graph actually shows :) it looks to be constant velocities between walking and riding home. Without knowing the route taken, the displacement would seem rather moot to me tho

  18. amistre64
    • 4 years ago
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    f(t) = 1/2 t + 9 ; t = [0,8) = 2t + c ; t = (8,10] havent the time to determine "c" tho

  19. phi
    • 4 years ago
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    The graph shows that at time=0 (call this the start time), you are 9 blocks from home. If you look carefully, at time= 3 minutes the graph crosses 8 blocks. That means you are walking 1 block every 3 minutes. At this rate (the slope of the line) it will take you 27 minutes to get home. But at time = 8.5 minutes you get in a car and get home in a total of 11 minutes. Notice that the car is much faster, and the slope of the line is much steeper.

  20. phi
    • 4 years ago
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    @amistre64 Your equations may confuse Creature. If you magnify the graph you see that the slope is 1/3 on the first line. The 2nd line goes from 8.5 to 11 mins

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