At time zero, a particle is at x = 4.0 m and y = 3.0 m and has velocity vector = (2.0 m/s)i + (-9.0 m/s)j. The acceleration of the particle is constant and is given by acceleration vector = (4.0 m/s^2)i + (3.0 m/s^2)j.
a) Find the velocity at t = 2.0 s
b) Express the position vector at t = 4.0 s in terms of i and j. In addition, give the magnitude and direction of the position vector at this time.
equations for velocity, position vector
The Attempt at a Solution
So far I have done a, which I think is correct.
velocity vector = (10 m/s)i - (6 m/s)j
(velocity of x is 10 m/s and velocity of y is -6 m/s)
I am stuck at b.
So far I have the displacement of x is 32 m. I got to that by:
x = 4 + 2 + (1/2)(4)(4^2) = 38 m
38 (final position) - 4 (initial position) = 32
For the displacement of y:
y = 3 - 9 + (1/2) (3)(4^2) = 18 m
so 18 - 3 = 15 m
so my final answer here would be:
position vector = 32i + 15j
is this correct? It seems like the equation for the position of y should involve gravity, but I have acceleration given.. please help!