posted by Christopher on .
Find the position and velocity of a particle at t=1.94 s if the particle is initially moving east at a speed of 19.8 m/s and experiences an acceleration of magnitude 4.08 m/s^2, directed west.
Magnitude and direction of the position.
direction- East of its original position
Magnitude and direction of the velocity.
direction- towards the east
I have figured out the directions for both of them, but I have been unable to figure out the magnitudes, I have tried many equations but none seem to work correctly. An answer with a detailed explanation is greatly appreciated.
V = 19.8m/s, East.
V = at=4.08m/s^2*1.94s = 7.92m/s,West.
Vn = 19.8 - 7.92 = 11.88m/s, East.
Vn = Net velocity.