Question

The position of a particle moving along the x-axis is given by: x=0.08sin(12t+.3)
meters, and t is measured in seconds. Determine the
position, velocity and acceleration of the particle when t=0.65 sec.

Answer 1

x=0.08sin(12t+0.3) position

v=x'=0,08cos(12t+0,3)*12
v=0,96cos(12t+0,3) velocity

a=v'=-0,96sin(12t+0,3)*12
a=-11,52sin(12t+0,3) aceleration

when t=0,65sg

x=0.08sin(12*0,65+0.3)=0,08sin(8,1)=0,08(0,9699)
x=0,0776 mts position

v=0,96cos(12*0,65+0,3)=0,96*cos(8,1)=0,96(-0,2435)
v=-0,2338m/s velocity

a=-11,52sin(12*0,65+0,3)=-11,52sin(8,1)=-11,52(0,9699)
a=-11,173m/s^2 aceleration