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?

The acceleration of a particle moving only on a horizontal xy plane is

1.	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?
Answer» SOLUTION :(a) `(72.0m)HATI+(90.7m)HATJ,` (B) `49.5^(@)`

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