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    Two forces F1 and F2 act on a particle P. The force F1 is given by F1 = (–i + 2j) N and F2 is a multiple of the vector (i + j). Given that the resultant of F1 and F2 acts in the direction of the vector (i + 3j), (a) find F2. The acceleration of P is (3i + 9j) m s–2. At time t = 0, the velocity of P is (3i – 22j) m s–1. (b) find the speed of P when t = 3 seconds.
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    (Original post by evlin)
    Two forces F1 and F2 act on a particle P. The force F1 is given by F1 = (–i + 2j) N and F2 is a multiple of the vector (i + j). Given that the resultant of F1 and F2 acts in the direction of the vector (i + 3j), (a) find F2. The acceleration of P is (3i + 9j) m s–2. At time t = 0, the velocity of P is (3i – 22j) m s–1. (b) find the speed of P when t = 3 seconds.
    Have a go at it yourself.
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    (Original post by username3234334)
    Two forces F1 and F2 act on a particle P. The force F1 is given by F1 = (–i + 2j) N and F2 is a multiple of the vector (i + j). Given that the resultant of F1 and F2 acts in the direction of the vector (i + 3j), (a) find F2. The acceleration of P is (3i + 9j) m s–2. At time t = 0, the velocity of P is (3i – 22j) m s–1. (b) find the speed of P when t = 3 seconds.
    F2 = c(i + j). Resultant = k(i + 3j).

    Resultant = F1 + F2
 
 
 
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