At 10:00 the boat is (8i-5j) km relative to O

At 12:50 the boat is at the point (-6i-12j) km relative to O

The motion of the boat is modelled as a particle moving at a constant speed on a straight line.

(a) Calculate the bearing on which the boat is moving.

For question (a), I 1. calculated the difference in position vectors 1. -6-8 i + -12+5 j = -14i -7j

I then calculated tan

^{-1}(14/7) = 64.434... deg

would this be correct? I am aware that this is not the velocity vector and that I would have to divide both magnitudes by the time elapsed between the two position vectors, that is 2 hr:50 mins = (17/6) hr = 2 .833333... hr

but surely it is just a scale factor so my answer angle would be the same?

(b) calculate the speed of the boat giving your answer in km hr

^{-1}

Finding the velocity vector

-(14/2.83333) i - (7/2.83333) j

2. I square root [ (14/2.83333)

^{2}+ (7/2.83333)

^{2}]

Are my workings and method correct?

Thank you,