A particle P moves in a horizontal plane. The acceleration of P is (–i + 2j) m s–2. At time t = 0, the velocity of P is (2i – 3j) m s–1. (a) Find, to the nearest degree, the angle between the vector j and the direction of motion of P when t = 0.
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Originally Posted by Aquafina A particle P moves in a horizontal plane. The acceleration of P is (–i + 2j) m s–2. At time t = 0, the velocity of P is (2i – 3j) m s–1. (a) Find, to the nearest degree, the angle between the vector j and the direction of motion of P when t = 0. direction of motion is degrees relative to the vector i so ... relative to the vector j , it would be ... what?
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