Question

I have a question about calculating the direction of the coast guard ship in Challenge Problem 1.3 in module 3. Is "direction" some sort of angle? Which angle would it be? I've calculated the direction of other vectors correctly in earlier problems, but the vector was in the first quadrant so it was easier. Here's the link to the problem and the solution: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/cartesian-coordinates-and-vectors/MIT8_01SC_problems03_soln.pdf. Thanks in advance!

Answer 1

Yea, If you treat the checkpoint as the origin of an xy coordinate plane, then the positions of the coast guard ship and the distressed sailboat can be plotted as vectors with their respective values and angles. The coast guard ship would be 42 degrees northwest form the checkpoint and would have a magnitude of 35 km. The sailboat would be 36 degrees southeast of the checkpoint and would have a magnitude of 20 km. What we are trying to find is the distance the coast guard ship would have to travel in order to reach the sailboat.

So you can think of this question as being: what vector must we add to the position vector of the coast guard ship (cgs) for it to be equal to the position vector of the sailboat (s)?
So:
R(cgs)+R(?)=R(s) we rearrange the equation and have R(?)=R(s)-R(cgs)
So to subtract these vectors, we must first decompose them into their components, using their relative angles and the pythagorean theorem.

X component of cgs vector=Rx(cgs)= -(35cos(42))=-26i km
its negative because I chose the convention of having increasing values of x to the right
*the i after the value just denotes that direction of the component is along the x axis

Y component of cgs vector=Ry(cgs)= 35sin(42)=23.4k km
positive because I chose y values to be increasing in the upward direction
*the k after the value just denotes that direction of the component is along the y axis

We repeat this decomposition for the sailboat position vector

X component of s vector=Rx(s)= 20cos(36)=16.2i km
its positive because I chose the convention of having increasing values of x to the right

Y component of s vector=Ry(s)= 20sin(36)=-11.8k km
negative because I chose y values to be increasing in the upward direction

now we subtract the vectors
R(?)=R(s)-R(cgs) where R(cgs)= -26i + 23.4k and R(s)= 16.2i - 11.8k
R(?)=(R(s)x-R(cgs)x)i + (R(s)y-(R(cgs)y)k
For this vector subtraction we must subtract the x and y components of the cgs vector from the x and y components of the s vector (that's the step you see above).
R(?)=(16.2-(-26))i + (-11.8-(23.4))k
R(?)=42.2kmi - 35.2kmk (which are the x and y components, respectively, of the resultant vector (the position vector that the coast guard ship needs to travel in order to reach the sail boat)

Now to find the angle, all we must do is use the tangent operation on the x and y components of the resultant vector (which all form a triangle) to find the angle that the coast guard ship must follow.
tan(angle)=(opposite/hypotenuse)
tan(angle)=(-35.2/42.2)
angle=inversetan(-35.2/42.2)
angle=-39.8 degrees

answer: about 40 degrees southeast
As you probably have figured out, the fact that the vectors belong to different quadrants (when you set the vectors up graphically) doesn't really complicate anything. You just have to make sure that your signs are correct all the way through.

Answer 2

I hope that helped. I didn't mean to spell things out too much, I just wanted to leave a reference for others who might not be as familiar with vectors. Here's the Diagram as well.

Answer 3

Thanks for the detailed solution. However, I'm still confused about which angle we are taking the tan of.... which vectors form the angle? and if we are using right triangle trigonometry, where is the right triangle? I appreciate your help.

Answer 4

Graphically, if we were to redraw, or shift, the resultant vector (the position vector the coast guard ship would have to travel) to the origin point (which we can do as long as we don't change the angle or magnitude of the vector), the angle that this vector makes with the x axis (which we consider to be east/west) would be the angle that the coast gurad ship would have to follow. Therefore your right triangle would consist of the x and y component of the resultant vector (the position vector the coast guard ship would have to travel) and the right angle formed between them.

We are able to do this (shift the resultant vector to the origin) because when we are calculating the angle at which that the coast guard ship would have to travel to reach the sailboat, the coast guard ship becomes the new frame of reference, and becomes our new point of origin from which we establish our cardinal directions. So essentially the answer would be app. 40 degrees southeast of the ship's current location.

I probably should've elaborated more on this in the first response....

I hope this answers your question :D
I also attached a diagram

Answer 5

OK that makes sense now! Thanks :)