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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.
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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marigirl
 one year ago
Best ResponseYou've already chosen the best response.01. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a=\frac{ d^2x }{ dt^2 }=8 \sin t\] 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=00+c1 c1=0 x=8 sint+7t \[when ~t=\frac{ \pi }{ 2 },x=8\sin \frac{ \pi }{ 2 }+7\frac{ \pi }{ 2 }=\frac{ 7 \pi }{ 2 }8\]
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