Question

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.

Answer 1

HI!! (again)

Answer 2

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

Answer 3

\[a=\frac{ d^2x }{ dt^2 }=8 \sin t\]
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
\[when ~t=\frac{ \pi }{ 2 },x=-8\sin \frac{ \pi }{ 2 }+7\frac{ \pi }{ 2 }=\frac{ 7 \pi }{ 2 }-8\]

Answer 4

thanks!

Answer 5

yw