## 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.

1. misty1212

HI!! (again)

2. marigirl

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

$a=\frac{ d^2x }{ dt^2 }=8 \sin t$ 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

thanks!

5. anonymous

yw