Question
A particle's constant acceleration is south at 2.1 m/s2. At t = 0, its velocity is 45.2 m/s east. What is its velocity at t = 7.7 s? What's the magnitude and direction?
Answers
drwls
This is a vector addition problem.
Vx = 45.2 m/s. It does not change.
Vy = (1/2) a t^2 = 1.05 t^2
Speed = sqrt(Vx^2 + Vy^2)
Direction (south of east) = arctan Vy/Vx
Crank out the numbers
Vx = 45.2 m/s. It does not change.
Vy = (1/2) a t^2 = 1.05 t^2
Speed = sqrt(Vx^2 + Vy^2)
Direction (south of east) = arctan Vy/Vx
Crank out the numbers