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marigirl

  • one year ago

Need clarification please for a integration question: Question: An object moves in a straight line so that after t seconds, its acceleration in ms^-2 is given as a=8sin t. If the velocity of the object after pi/2 seconds is 7 ms^-1, find its distance s, from its initial position at that movement.

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  1. misty1212
    • one year ago
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    HI!! (again)

  2. marigirl
    • one year ago
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    1. Integrate acceleration to obtain velocity v= -4cos 2t+c and obtain c by using t=pi/2 and 7ms^-1 and now v=-4cos 2t+3 2. integrate velocity formula to obtain distance formula S=-2sin 2t+ 3t +c this is where i got a bit confused, can we regard c as zero since this is the initial movement? and continue as S=-2sin 2t+ 3t and find the distance when t=pi/2

  3. anonymous
    • one year ago
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    \[a=\frac{ d^2x }{ dt^2 }=8 \sin t\] int. w.r.t.t \[\frac{ dx }{ dt }=-8 \cos t+c\] when \[t=\frac{ \pi }{ 2 },v=7 m/s\] \[v=\frac{ dx }{ dt }\] \[7=-8 \cos \frac{ \pi }{ 2 }+c\] 7=-8*0+c,c=7 \[\frac{ dx }{ dt }=-8\cos t+7\] integrate again w.r.t. t \[x=-8 \sin t+7t+c1\] when t=0,x=0 0=0-0+c1 c1=0 x=-8 sint+7t \[when ~t=\frac{ \pi }{ 2 },x=-8\sin \frac{ \pi }{ 2 }+7\frac{ \pi }{ 2 }=\frac{ 7 \pi }{ 2 }-8\]

  4. marigirl
    • one year ago
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    thanks!

  5. anonymous
    • one year ago
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    yw

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