physics
posted by Need Help on .
A particle experiences a constant acceleration that is south at 2.50 meters per second squared. At t=0, its velocity is 40.0 meters per second east. What is its velocity at t=8.00 seconds?

The acceleration is due south, so the eastwest component (40 m/s) is not altered.
For the northsouth component, at t=0,
initial velocity = 0 m/s
after 8 seconds,
velocity v=0+(2.5)8 = 20 m/s (<0 for south)
Velocity = (40,20) m/s
Magnitude = √(40²+20²)
=44.72 m/s
direction:
atan(20/40)=26.57°, or
26.57° south of east.