1. Oct 8, 2006

### ch3570r

I posted this in the Intro Physics section, but found now help, so hopefully one of you can help me out.

Here are the problems:

1. A football player runs directly down the field for 35 m before turning to the right at an angle of 25degrees from his original direction and running an additional 15 m before getting tackled. What is the magnitude and direction of the runner's total displacement? (answers are 49m, and 7.3degrees, but how do you get them??)

2. A plane travels 2.5 km at an angle of 35degrees to the ground, then changes direction and travels 5.2 km at an angle of 22degrees to the ground. What is the magnitude and direction of the plane's total displacement?? (I dont know the answer to this one)

I've tried over and over to figure it out, usually attempting to split the diagram into its X and Y components, but I cant seem to figure them out

Any help would be very appreciated.

Last edited: Mar 22, 2010
2. Oct 8, 2006

### OlderDan

Finding components to add is the right approach. Tell us what you did for #1 and we can see where you went wrong.

3. Oct 9, 2006

### ch3570r

The first I drew was the 35 m displacement vector, which is pointing straight up (vertical). The next thing I drew is the vector of the player running 15 m at a 25 degree angle to the direction he was initially running. At this point I broke up the 15 m segment into its component (x and y) vectors. Using trigonometry, I found that the second segment of the run was equivalent to him running 15*cos(25) m up and 15*sin(25) m to the right. To find the total displacement, I used the Pythagorean theorem: displacement = sqrt( (35 + 15cos(25))^2 + (15sin(25))^2 ). This comes out to 49 m.
BUT, I thought I was supposed to get the square root of 49, not leave it as is. Atleast, that gives me the answer to the first part.

As for the angle, I tried finding the inverse tangent of the numbers I got above, but it did not come out to 7.3 degrees. Anyone know why, or if Im doing it right?

4. Oct 9, 2006

### OlderDan

You are doing the first part right. You are already taking the square root of the sum of squares to get 49. You don't want to take the square root a second time. The two terms you squared to find the length of the resultant are corect. The ratio of those terms (not their squares) is the tangent of the angle you are looking for. I get 7.43 degrees rather than 7.3, but that's close.

5. Oct 9, 2006

### ch3570r

Thanks Dan

Ok. So my tangent angle is equal to (35 + 15cos(25) / (15sin(25) ? I get about 7.6....That seems too far off. I thought that when finding an angle inside of a triangle, I had to use the inverse tangent, which would give me something too far off (in the 80's).

On the second problem, I saw that they gave me 2 angles instead on 1. Im guessing this means I need two separate triangles instead of one like the first ? Well, I went ahead and separated them, and then combined the two X and Y components of the triangles. How do I handle the angles of the two triangles, by that I mean once I add the vectors of each triangle, do I keep the original angles? (22 and 35)

And, if I leave the original angles the same, would that make my new angle 123degrees (180 - (22+35))??

6. Oct 9, 2006

### OlderDan

You do want the inverse tangent, but you have to get the length of the opposite side in the numerator and that of the adjacent side in the denominator. That angle in the 80s is the complement of the angle you are looking for.

The only difference for problem 2 is that you have fo find components for both of the vectors you are adding instead of just the one. Everything else is the same.

7. Oct 9, 2006

### ch3570r

Thanks again for all the help. I found the angle by doing finding tan-1(35 + 15cos(25) / (15sin(25), then subtracting that from 90, sense they're complimentary angles.

I also tried the second problem, and my final answers came out to be 3.68Km, and 29.8degrees SW. Im not sure if its right, but we'll see tomorrow.

Thanks once more for the help

8. Oct 9, 2006

### OlderDan

3.68Km cannot be right. It has to be more than either of the two vectors you are adding, unless the plane turns around and heads back the way it came as well as changing its climb angle. Since nothing was said about turning around, I doubt they intended that.

From your answer it appears you are taking the angles to mean directions on a map. To me it sounds like they are talking about elevation angles, although it would take a high performance plane to achieve those. If you have a diagram that I don't see, maybe I am interpreting it wrong.

Last edited: Oct 9, 2006
9. Oct 9, 2006

### ch3570r

it doesnt give me a diagram, only the word problem that I gave above:

2. A plane travels 2.5 km at an angle of 35degrees to the ground, then changes direction and travels 5.2 km at an angle of 22degrees to the ground. What is the magnitude and direction of the plane's total displacement?? (I dont know the answer to this one)

it does say it "changes direction", but without a diagram, I too am lost.

10. Oct 9, 2006

### OlderDan

I would then assume the plane maintains a constant heading. The 35 degrees and 22 degrees are angles of climb, not horizontal directions. You can draw the two vectors on a plane with the x-axis in the direction of the heading and y-axis vertical and go from there.