A particle's constant acceleration is south at 2.1 m/s2. At t = 0, its velocity is 42.6 m/s east. What is the magnitude and direction of its velocity at t = 7.6 s?
Use the formulas
Y (south) = (a/2) t^2
Vy = at
X (east) = Vo * t
Vx = Vo
where Vo = 42.6 m/s and a = 2.1 m^2
At any timne t,
speed = sqrt [Vx^2 + Vy^2}
direction (south of east)
= tan^-1 (Vy/Vx) = tan^-1 (a t/Vo)