- #1
Kernul
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Homework Statement
The acceleration of a particle moving only on a horizontal xy plane is given by a = 3ti + 4tj, where a is in meters per second squared and t is in seconds. At t = 0, the position vector r = (20.0 m)i + (40.0 m)j locates the particle, which then has the velocity vector v = (5.00m/s)i + (2.00m/s)j. At t = 4.00s, what are
(a) its position vector in unit-vector notation and
(b) the angle between its direction of travel and the positive direction of the x axis?
Homework Equations
x = x0 + vox * t + 1/2 * ax * t2
y = y0 + voy * t + 1/2 * ay * t2
The Attempt at a Solution
So, since ax = 3t and ay = 4t, I got the following values for the position vector:
(136m)i + (176m)j
The problem is that the values aren't the solution given by the book (which is (72m)i + (90.7m)j ).
I keep reading but nothing comes to my mind besides the fact that probably I did something wrong with the acceleration.