The acceleration of a particle moving only on a horizontal xy plane is given by a→=4ti^+5tj^, where a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector r→=(28.0m)i^+(48.0m)j^ locates the paticle, which then has the velocity vector v→=(6.70m/s)i^+(2.30m/s)j^. At t = 4.60 s, 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?

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A: v→ = - 20 j^ m/s a→ = 1 m/s2 i^

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