Physics problem (vectors)?

Question

A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.40 km northeast, then an additional distance in an unknown direction. Her final position is 6.6km directly east of the starting point. Find the magnitude and direction of the third leg of the voyage.

Can anyone help me with this? I'm not sure how to do this and my book gives a very poor explanation.

Answer 1

Here's how to go about solving this problem. Draw the arrows to signify vectors.

..................^

................ /   \

................ /      \

Start ------> _____ END

The horizontal distance between start and end must be 6.6 km. The upward sloping distance is 3.4 km (directed at a 45 degree angle). This means the 3.4 km gives us a horizontal distance of 3.4 cos45 = 2.4 km. Adding that to the original 2 km East, we have 2+2.4 = 4.4 km. Thus the final vector needs to have a horizontal component of 6.6-4.4 = 2.2 km. As for the vertical component, it must be the same as the upward sloping vertical component, which is 3.4sin45 = 2.4 km. Thus we know the vertical and horizontal components are 2.4, and 2.2 km respectively.

To figure out the magnitude, simply take the square root of the sum of the squares, or SQROOT ( 2.4^2 + 2.2^2) = 3.25 km. To find the angle, take the arc tangent of (2.4 / 2.2) = 47.49 degrees (above the horizontal line).

Answer 2

The journey looks like a straight line east 2 km then turns north 45 degrees and sails 3.4 km then turns southeast and sails back to the eastward line ( the bearing) she was originally on

  Since she eneded 6.6 km E of the start and she had initially sailed 2 km E then 6.6 - 2 =4.6 is the base of the triangle with one side 3.4 km and included angle 45 degrees. Now you have to find the opposite side of that triangle using trig. I'd use the Law of Cosines. Then I'd use the Law of Sines to find the angle in the triangle at the second point in her voyage. Her start point is 0. point 1 is 2km

Using the Law of Cosines

I get the third side of the triangle to be 3.26 km

Using the Law of Sines

I get the angle at point 2 in her journey

Angle = 85.5 degrees in the triangle. So she must turn 180 -85.5 = 94.5 degrees Southeast

Since she was already 45 degrees east of north adding 94.5 degrees southeast makes her bearing 45 +94.5 = 139.5 degees east of north