Maths: Vector Questions?

Question

Here i have two question that i do not understand about the questions:
1) A yacht sails 15 km on a bearing 248° followed by 17 km on a bearing 297°. Calculate the distance (in km to the nearest 0.1 km) and bearing (to the nearest degree) of the yacht's final position from its initial position.

2) A particle has an initial position vector of (-7 i + 15 j) m with respect to an origin O. If the particle moves with a constant velocity of (5 i - 8 j) m/s, what will be its position vector after 14 seconds?

If can give the details to show the solution for each question. Thanks.

Answer 1

Question 2 is easier. Do the i's and j's separately. For i, you have -7 + 14*5 = -7 + 70 = 63. For j, you have +15 - 14*8 = +15 - 112 = -97.

After 14 seconds, the position vector will be 63 i - 97 j.

Since I'm not a sailor, I don't understand bearings. I'll assume 0 degrees and 360 are north; 90 is east; 180 south; and 270 west. That means 248 degrees is in the third quadrant (SE), 22 degrees south of west; and 297 degrees is in the second (NW) quadrant, 27 degrees north of west.

When you draw and add these vectors, you can see that the yacht is generally heading in a westerly direction. Let's do the x (westerly) component first, using cosine. His x-component is 15 cos 22 + 17 cos 27 = 29.055 km W.

For the y (N-S) component, we must subtract sines, since one vector is south, and the other north. Also, the north component will be greater, so we get 17 sin 27 - 15 sin 22 = 2.099 km N.

The Pythagorean Theorem will give the final distance: sqrt( 29.055^2 + 2.099^2) = sqrt(848.59) = 29.13 km ==> 29.1 km.

For the bearing, get the tangent of the angle relative to 270 degrees. This tangent is 2.099 / 29.055 = 0.07224, so the angle is 4.13 degrees north of west, with a bearing of 274 degrees.

So your answer is 29.1 km, 274 degrees.

Answer 2

Firstly, I would recommend that posters only give you hints and not a worked through solution - you will not learn anything that way.

I recommend to get some graph paper and a protractor to draw out paths of the yacht.

Draw two axes on the paper as a cross and label the centre O, the axis going upwards as N(north), then E,S,W for the other three axis directions.

The yacht first travels in a general south-westerly direction for 15km and then turns north-westerly and travels for a further 17km.

It actuall goes 22 degrees south of due west for 15km and then 27 degrees north of due west for 17km. So the sum of these two paths with be a little bit north of due west from the origin O.

If you draw these two paths and another path from O to the destination you will have a triangle.

The easiest way to calculate the angle of the last line from O to the destination and to calculate the distance of that line is to "resolve" the two original vectors into vertical and horizontal components. You then add the two horizontal and two vertical components together to get a horizontal and vertical component of the vector that you are looking for.

The length of the vector will be the hypotenuse of the right-handed triangle formed by the horizontal and vertical vector sum components.

The angle from the westerly horizontal will be the inverse tangent of the vertical component/horizontal component plus 270 degrees to get you in the right quadrant of the graph from O (due north being zero degrees).

The particle question is even easier because you are already given the horizontal and vertical components of the particle position. In this case just draw cross axes on the graph paper and lable them as before N->i, E->j, S->-i, W->-j

Count squares in direction -i and j for the first vector to get the start position. From that position count squares in directions i, j for the second vector to get a new position. Now repeat the second step from this new position 13 times to give a total of 14 second at that speed.

Again you need to add up all the horizontal and vertical components. In this case you don't need to calculate the hypotenuse because the question asks for the final position which will be ?i, ?j

The ?s will be the sums of horizontal (j) and vertical (i) components which is the total number of squares you count in i and j directions.

Answer 3

1) use the cosine law
http://ilearn.senecac.on.ca/learningobjects/MathConcepts/CosineLaw/main.htm

You know b,c and A. Solve for a.
It helps to draw it.

Now that you know a, you can use the sine law
http://ilearn.senecac.on.ca/learningobjects/MathConcepts/SineLaw/main.htm
To solve for the angle off of the original course.

2) break it into components. If I understand correctly (and it has been a while) you would move 14 times the vector. Just add the i and j components. So i position would be -7+14*5 j would be 15-8*14.