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AravindG

  • 5 years ago

how to calculate v,and a from this graph??

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  1. AravindG
    • 5 years ago
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    |dw:1327254348794:dw|

  2. AravindG
    • 5 years ago
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    u see i need a and v at -1.5

  3. anonymous
    • 5 years ago
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    Put some labels on the axes :D.

  4. AravindG
    • 5 years ago
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    actually i need only sign of a and v

  5. anonymous
    • 5 years ago
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    Like gogind said, we need labels. We have no idea how "v" or "a" relates to this sine wave.

  6. AravindG
    • 5 years ago
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    |dw:1327295200296:dw|

  7. AravindG
    • 5 years ago
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    sorry for tht

  8. anonymous
    • 5 years ago
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    If that is x on the y-axis, as in distance. then you would get v(t) as at derivative of x(t) with respect to t (change in distance over time is velocity) . Similarly you would get a(t) as the derivative of v(t) with respect to t or second derivative of x(t) with respect to t (change is velocity over time is acceleration). Now you have to figure out what x(t) is. It is \[\ x(t) = -sin(\pi t)\] so now: \[\ v(t) = \frac{d x(t)}{dt} \]\[\ a(t)=\frac{dv(t)}{dt}=\frac{d^2x(t)}{dt^2}\] All you have to do now is differentiate.

  9. anonymous
    • 5 years ago
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    Actually, \[\ x(t) = -Asin(\pi t)\], Where A is the amplitude of the sine wave. You didn't provide the amplitude so I assumed it's 1.

  10. anonymous
    • 5 years ago
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    |dw:1327325940478:dw| always draw a tangent to the given point at the curve here at -1.5m,tangent6 is parallel to x-axis so NO ANGLE MADE WITH X-AXIS now for acceleration,take any point b before the time (-2 seconds is less than -1.5 seconds) then draw a tangent at b we cxan see that it makes an obtuse angle with x=axis meaning -ve velocity accleration=0-(-velocity)/time=+ve acceleration edhe pole tangent varachu nooki angle nooki sum cheidhal answer kitum!

  11. AravindG
    • 5 years ago
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    thx salini i was looking for tht answr very helpful thx thx gogind too

  12. AravindG
    • 5 years ago
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    one doubt y we take v as -ve wen angle is obtuse?

  13. anonymous
    • 5 years ago
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    the basic trignometry defines tan@ of obtuse angles to be -ve(2nd quadrant)

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