The acceleration of a particle moving only on a horizontal xy plane is given by `veca=3thati+4thatj`, where `veca` is in meters per secondsquared and t is in seconds. At t = 0, the position vector `vecr=(20.0m)hati+(40.0m)hatj` locates the particle, which then has velocity vector `vecv=(5.00m//s)hati+(2.00m//s)hatj`. At t = 4.00 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the position direction of the x axis?

Explanation

(a) `(72.0m)hati+(90.7m)hatj,` (b) `49.5^(@)`