Basic Physics (Vector)

Question

Please help me I dont get this one, also provide a complete solution with final answer and in depth step-by-step procedure. Include also a clear graph.

a.) = + + +

= 25 mi, north

= 15 mi, 45 degrees north of west

= 8 mi, 30 degrees south of west

= 11 mi, south

Thank you so much! I'm so having a hard time getting the idea on how to do this one.

Answer 1

There are two ways to add vectors. One is to start at (0,0) and draw all of them to scale with a ruler and protractor. (You could make one mile equal 1 cm or whatever.) Where the first vector ends, the next one begins. We call this head-to-tail. If you draw them all accurately, then you can draw a final vector that goes from (0,0) to the end point of vector D. The length, angle, and direction of this final vector is the answer. You're basically simulating the exact path the plane traveled by drawing that path on a piece of paper.

That method is super tedious and inaccurate unless your equipment is good. A much better method is to use the component method and just work out the numbers without drawing anything!

Any action happening at an angle is equivalent to two separate actions combined that DIDN'T happen at an angle. For example, walking 10 miles northeast has the same outcome as someone first walking 8 miles directly east, then 6 miles directly north. (10² = 8² + 6²). The person who walked 8 miles then 6 miles definitely took more steps and burned more calories than the person who took the direct 10 mile route. But the RESULT of their walking... where they ended up... was the same. If you goal is to reach that spot, it doesn't matter if you go in a straight line (like a helicopter) or on the east freeway then the north freeway (like a car). They both reached the destination!

On to your question. So Vector A and vector D are easy because they are not at an angle. Nice and simple movement. But for vector B, the plane is flying both west AND north! It is much simpler if we deal with both of those separately by chopping them up with some trig.

15 * sin(45) = 10.6 miles. 15 * cos(45) = 10.6 miles. So instead of saying the plane flew 15 miles northwest, let's pretend it first went 10.6 miles north then 10.6 miles west.

For vector C, we chop it up into how far west, and how far south it went.

8 * sin(30) = 4 miles. 8*cos(30) = 6.9 miles. Since the movement is 30 degrees south OF west, then the plane's motion was more towards the west than the south. So it went 6.9 miles west (the bigger number) and 4 miles south (the smaller number).

Okay. We started with two simple movements (north and south) and two complicated movements (northwest and southwest). 4 Vectors in total. But we can manage everything way better by pretending it was 6 simple motions that only exist going exactly north, south, east, or west.

So where did the plane end up? Well, for north/south it went 25 north, 11 south, 4 south, 10.6 north. 25 + 10.6 - 11 - 4 = 20.6 miles north. For east/west, it went 6.9 west, 10.6 west. = 17.8 miles west.

So the plane ended up 20.6 miles north of where it started, and 17.8 miles west of where it started. (It took a very weird path to get there, but that's where it ended up!) If you draw a triangle with one arrow going 17.8 west and the other going 20.6 north, the hypotenuse of that triangle would be sqrt (17.8² + 20.6²) = 27.2 miles northwest. That is how far a direct route would need to be if you wanted to get the same outcome as this plane without using so much wasted fuel wandering back and forth.

The direction of vector R is northwest, but at what angle is it? Well, trig is normally used when you know the angle of a triangle but are curious about one of its sides. In this case, we know the sides of the triangle and want to know the angle! This calls for inverse trig. Since we have all 3 sides of the triangle, we can use inverse sin, inverse cos, or inverse tan and any of them would work. tan^-1(17.8/20.6) = 40.8 degrees. But is that north of west, or west of north? Well the plane ended up more north than west, and 40.8 degrees west of north would make that true. You could also say it was 49.2 degrees north of west and that would also be true, since they add to 90 degrees.

Vector R is 27.2 miles, heading 40.8 degrees west of north. Next time I hope the plane only flies north/south/east/west to save you from all this extra work chopping vectors up into components! =D

Answer 2

Renan, I can show you how to get the answer, and you can get the actual numbers and angles.

first, what is the solution like? It is to be a vector. That is, it will have a length (in miles) and an angular direction ( in degrees).

The step-by-step instructions will consist of you writing something like: “Vector A (or B, C , or D) begins at location (x, y), and moves in direction alpha (or beta, gamma ,or delta) for a distance of m miles. We can compute the endpoint of this vector from the distance and angle, yielding this endpoint: “ and you’ll put the computed endpoint.

you’ll do that four times, once for each vector, starting where the previous endpoint. When you’re done, you’ll have a final endpoint, and you’ll know its (x,y) coordinates. From those you’ll compute the distance and the angle for the whole string of vectors.

Let’s start.

start at (0,0). Vector A goes North 25 miles. Since we usually model “North” as positive Y values, the endpoint is simply (0, 25). That was easy, but it gets more involved.

Vector B starts at (0,25) and moves 15 miles in the direction 45 degrees north of west. So from (0,25) draw a dotted line straight West. That means in the direction of negative X. But that’s not where the vector goes. It is a solid line going at a 45 degree angle upward from that dotted line. Since the length of vector B is 15 miles, you can compute the endpoint’s X value by using the cosine of 45.

Adjacent / hypotenuse = cos(45), solving for adjacent, we get

adjacent = hypotenuse * cos(45) = 15*sqrt(2)/2. Since this is the change in X, and it is the negative direction,

X(B) = X(A) - 15sqrt(2)/2 = (0-(15/2)sqrt(2))

(I would leave it in those terms until the end; it’s easier to do math with square roots than with numbers themselves. )

compute the Y value by using

opposite / hypotenuse = sin(45), solving for opposite, we get

opposite = hypotenuse *sin(45) =

Y(B) = Y(A) + 15sin(45) = Y(A)+ 15sqrt(2)/2

= Y(B) = 25 + (15sqrt(2)/2)

now we start vector C where B left off. That is, C’s starting point is B’s endpoint.

Vector C moves in the direction 30 degrees south of West. Once again, draw a dotted line moving to the left from

( -(15/2)sqrt(2), 25 + (15/2)sqrt(2) ), and then draw a dotted line at an angle 30 downward from that dotted line. Since the distance is the hypotenuse of 8 miles, this will cause changes to X and Y as follows:

X(C) = X(B) - 8cos(30) = X(B) - 8sqrt(3)/2 = X(B) - 4sqrt(3)

= (-15/2)sqrt(2) - 4sqrt(3)

and Y(C) = Y(B) - 8Sin(30) = Y(B) - 8*1/2 = Y(B) -4

= (25 + (15/2)sqrt(2)) -4) = (21+ (15/2)sqrt(2))

finally the fourth vector goes South, the direction of negative Y.

so draw a line straight down from (X(C), Y(C)) a distance of 11 miles. This makes easy changes to X and Y.

Y(D) = Y(C) - 11 = 21+ (15/2)Sqrt(2) - 11

= 10 + (15/2)sqrt(2))

and X(D) = X(C)

so the resulting vector has the endpoint

( (-(15/2)sqrt(2)) - 4sqrt(3), 10 + (15/2)sqrt(2)))

and it begins at (0,0).

now you can evaluate those square roots to find the (x, y) coordinates of the endpoint and from them, the length of the vector (by the Pythagorean theorem). The angle is the the inverse tangent of X(D)/Y(D).