mechanics (vectors)

• Apr 29th 2009, 07:41 AM
Aquafina
mechanics (vectors)
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.
• Apr 30th 2009, 05:59 AM
skeeter
Quote:

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 $\theta = \arctan\left(-\frac{3}{2}\right) = -56.3$ degrees relative to the vector i

so ... relative to the vector j , it would be ... what?