Question

At time t in seconds, a particle’s distance s⁢(t), in centimeters, from a point is given by s⁢(t)=13sin⁢ t+40. What is the average velocity of the particle from t⁢= π3 to t⁢= 13π3?

Answer 1

Step-by-step explanation:

Given

distance
`s(t)=13\sin (t)+40`

at
`t=(\pi )/(3)`

`s((\pi )/(3))=13\sin ((\pi )/(3))+40`

`s((\pi )/(3))=13* (√(3))/(2)+40`

at
`t=(13\pi)/(3)=4\pi+(\pi)/(3)`

`s(4\pi+(\pi)/(3))=13\sin (4\pi+(\pi)/(3))+40`

Average velocity is change in position w.r.t time

`v_(avg)=(s(4\pi+(\pi)/(3))-s((\pi )/(3)))/(4\pi+(\pi)/(3)-(\pi )/(3))`

`v_(avg)=0`