Year 12s: what will be the first way you research your future options? >>

M1 Vectors

A model boat A, moves on a lake with constant velocity (-i+6j)ms^-1. At time t=0, A is at the point with position vector (2i-10j)m. Find

a) speed of A = 6.08

b) the direction in which A is moving as a bearing = 351

c) a time t=0, a second boat B is at the point with position vector (-26i+4j)m, given that the velocity of B is (3i+4j)ms^-1.

Show that A and B will collide at Point P, and find the Velocity Vector P = (-5i+32j)

and finally d) the part I'm stuck on

Given instead that B has speed 8ms^-1 and moves in the direction of vector (3i+4j)

Find the distance of B from P when t=7s

Answer Workings: New velocity of B = 8/5(3i+4j)ms^-1

I dont understand where this 1/5 has come from? the 8 is the speed, but what's 1/5?

final answer = 21

Reply 1

insparato

The unit vector of 3i + 4j is 3i + 4j/5.

So the Velocity is 8/5 (3i + 4j).

I think!

Reply 2

hopinmad

I think it's saying that the velocity is "speed times unit-vector-in-direction-of-travel". The unit vector is 3i+4j divided by the size, or magnitude of the vector, i.e. sqrt(3^2+4^2) so that you get 8 x (3i+4j)/5.