Physics question:Particle leaves the origin

[Judas543]

Homework Statement

A particle leaves the origin with initial velocity v0 = 11i+ 14 j m/s. It undergoes a constant acceleration given by a = –1.2 i + 0.26 j m/s2


(a) When [i.e., for t greater than zero] does the particle cross the y axis?
t = s
(b) What is its y coordinate at that time?
y = m



(c) How fast is it moving, and in what direction, at that time?
v = m/s


θv = ° from the +x axis

Homework Equations

Integrals Calculus I believe

The Attempt at a Solution

v(t) = ∫ a dt + C = -1.2 t i + 0.26 t j + 11 i + 14 j
x(t) = ∫ vx dt + C1 = -0.6 t^2 + 11 t
y(t) = ∫ vy dt + C2 = 0.13 t^2 + 14 t

a) t=18s
b) y= 300m
c)= 22m/s
d) Can't figure out... What direction is it moving at that time? ____ degrees

 

[Zhivago]

for d) , you have the x,y components of the velocity vector, and you know that tan(teta) = y/x
Just imagine it as a triangle.

Hope this helps