The acceleration of a particle moving only on a horizontal xy plane is given by a=5ti+6tj, where a is in meters per second-squared and t is in seconds. At t=0, the position vector r=(19.0m)i+(39.0m)j locates the particle, which then has the velocity vector v=(5.70m/s)i+(3.40m/s)j.At t=3.60s, 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?
For a) I got the answer -4.11 m/s. (This is correct)
I am having trouble finding b) though.
Physics - Anonymous, Saturday, September 12, 2015 at 10:24pm