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saifoo.khan

  • 4 years ago

Quick Question, Area under speed-time graph is the distance travelled?? T or F?

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  1. anonymous
    • 4 years ago
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    t

  2. dumbcow
    • 4 years ago
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    True speed = distance/time *time = distance

  3. anonymous
    • 4 years ago
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    umm is this a physics question?

  4. saifoo.khan
    • 4 years ago
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    No, this is Math,

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  5. anonymous
    • 4 years ago
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    okies :)

  6. saifoo.khan
    • 4 years ago
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    Now it's fine. :D

  7. anonymous
    • 4 years ago
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    Distance to me is an absolute value. No directional sense. So, as I pointed in the link and as MrYantho suggested, I like displacement.

  8. saifoo.khan
    • 4 years ago
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    Okay!! (: can u solve the question in the picture?

  9. anonymous
    • 4 years ago
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    Saifoo - sorry man....I did this stuff 22 years ago and it is hard to remember. I can google it up and get it. But, you or someone current is going to be a lot better at it.

  10. saifoo.khan
    • 4 years ago
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    No worries. Thanks for trying! :)

  11. anonymous
    • 4 years ago
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    sorries me cant do it either :(

  12. saifoo.khan
    • 4 years ago
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    np

  13. anonymous
    • 4 years ago
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    um its just at constant speed not instantaneous right?

  14. saifoo.khan
    • 4 years ago
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    first it has increasing speed, then constant speed, i want to draw the distance-time graph for this.

  15. anonymous
    • 4 years ago
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    could you wait a bit ill try to help this stuff is in my physics book

  16. anonymous
    • 4 years ago
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    its the math it uses

  17. saifoo.khan
    • 4 years ago
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    This might help,

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  18. anonymous
    • 4 years ago
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    the formula you just use is a regular d=r/t

  19. saifoo.khan
    • 4 years ago
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    idk. this is the screenshot form the marking scheme.. i have no idea how they did that.

  20. anonymous
    • 4 years ago
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    formula for aceleration final speed= initial speed + speed change

  21. saifoo.khan
    • 4 years ago
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    \[a = \frac{v-u}{t}\]Right?

  22. saifoo.khan
    • 4 years ago
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    But how we will draw the speed-time? :l

  23. anonymous
    • 4 years ago
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    on you graph y is the speed x is the time

  24. saifoo.khan
    • 4 years ago
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    wait. sorry.. that is already drawn in the question.. i want displacement time. hehe,

  25. anonymous
    • 4 years ago
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    displacement ? umm i dont understand displacement from what?

  26. saifoo.khan
    • 4 years ago
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    read the question again, that might help you the part (i) http://assets.openstudy.com/updates/attachments/4f1a4dd4e4b04992dd22a68d-saifoo.khan-1327124195861-dscn4553.jpg

  27. anonymous
    • 4 years ago
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    it is cut off

  28. anonymous
    • 4 years ago
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    can you type the rest it?

  29. saifoo.khan
    • 4 years ago
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    The first part says to sketch the region whose area represents and stuff.. t And the cut off part says "find the value of T"

  30. anonymous
    • 4 years ago
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    first time change of 1 cyclist s=(v) (t) + 1/2 ( a)( t) use formula again to find displacement again of the cyclist

  31. anonymous
    • 4 years ago
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    then you may probably use d=rt to figure out the rest of the stuff i think this is how it is done :\

  32. anonymous
    • 4 years ago
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    it's true that the area under speed-time graph is distance you can get this from this formula d=vt |dw:1327401093475:dw|

  33. dumbcow
    • 4 years ago
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    The line for cyclist P is v = 2t, since the constant acceleration is given as 2 the displacement is 16 and the area under the curve is the displacement Area = vt/2 = 16 vt = 32 substitute v=2t 2t^2 = 32 t^2 = 16 t =4 Therefore, cyclist Q starts 4 sec after when cyclist P is going 8 m/s

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