Question

A particle moving initially at 4 m/s in the positive x direction is given an acceleration of 3 m/s^2 in the positive y direction for 2 seconds.

What is the magnitude and direction of the particle's velocity after 2 seconds?

If the particle was located at the origin when the acceleration began, what are its x and y coordinates after 2 seconds?

Answer 1

V = a * t = 3 m/s^2 * 2 s = 6 m/s in
positive y-direction,
V = 4 + i6,
tanA = Y/X = 6/4 = 1.5,
A = 56.3 Deg.,
MAG. = X/cosA = 4/cos = 7.2 m,
V = 4 + i6 = 7.2 m @ 56.3 Deg.,

(4 , 6).