A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 4.5 m, y = 6.5 m, and has velocity v(naught) =
(6.5 m/s)i-hat + (−2.5 m/s)j-hat (arrow for v(naught) points toward positive x direction). The acceleration
is given by a = (6.5 m/s2)i-hat + (1 m/s2) j-hat (arrow for a points in positive x direction).
1.) What is the x component of velocity after 9 s?
2.) What is the y component of velocity after 9 s?
How do I get this? We went over one example in class and I have 2 formulas to use:
1.) v(subx)=v(naughtx) or (vnaught)x (just realized I took poor notes)
2.) x=v(naughtx)t or x=v(naught)xt
Our class example didn't use i or j hat and didn't ask for x or y after a given time.
Answers given are 65 m/s for x and 6.5 m/s for y; need to figure out how to get them.
website for this question is webphysics.iupui.edu/152/Hw/HW04.pdf
questions #1-3
V=Vi+at is a vector equation
1) Vx=Vix+ ax*t
=6.5m/s + 6.5m/s^2*9=65m/s
do the same with the y component