m1 velocity vectors

A particle P moves with a constant velocity (3i+2j)ms-1 with respect to a fixed origin O. It passes through the point A whose position vector is (2i+11j)m at t=0.

a) find the angle in degrees that the velocity vector of P makes with the vector i. (2)

the answer is tan^-1(2/3) = 33.7 degrees

can someone explain how they did this, and diagrammatically if possible.

b)calculate the distance of P from O when t = 2 (4)

the answer is 17, but how

thank you

Reply 1

Malik

12

ash213

A particle P moves with a constant velocity (3i+2j)ms-1 with respect to a fixed origin O. It passes through the point A whose position vector is (2i+11j)m at t=0.

a) find the angle in degrees that the velocity vector of P makes with the vector i. (2)

the answer is tan^-1(2/3) = 33.7 degrees
thank you

Its really simple P(3i+2j) ie 3units to the right 2 units up. Join them up and you have right angled triangle. With opposite=2 adjacent=3
.:. tan x= 2/3

Reply 2

asanonymous

initial position vector of P is (2i + 11j)

displacement = velocity x time, so the position vector of P from o at t seconds is (2i + 11j) + (3i + 2j)t

sub t=2 therefore (2i + 11j)+ (6i + 4j) which equals (8i + 15j)

magnitude of this is shown by square root of (8 squared + 15 squared) = square root of 289 = 17

m1 velocity vectors

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