Adding vectors algebraically for football runner

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Homework Statement



A football player runs directly down the field for 35m before turning to the right at an angle of 25 degrees from his original direction and running an additional 15m before getting tackled. What is the magnitude and direction of the runner's total displacement?


Homework Equations


None that I know of.

The Attempt at a Solution


[PLAIN]http://img269.imageshack.us/img269/7127/whatigotv.png [Broken]

This is how far I get, its not a right triangle so I am pretty lost :/
 
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Answers and Replies

  • #2
berkeman
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Homework Statement



A football player runs directly down the field for 35m before turning to the right at an angle of 25 degrees from his original direction and running an additional 15m before getting tackled. What is the magnitude and direction of the runner's total displacement?


Homework Equations


None that I know of.

The Attempt at a Solution


[PLAIN]http://img269.imageshack.us/img269/7127/whatigotv.png [Broken]

This is how far I get, its not a right triangle so I am pretty lost :/

Welcome to the PF!

Convert the vectors into rectangular coordinates, add the x components, add the y components, and convert back to polar form. You will generally always add vectors in rectangular coordinates, so get good at converting back and forth.

Show us what your calculation looks like now.

(And BTW, try to post smaller images in the future so they don't blow up the size of the page. Thanks.)
 
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  • #3
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Welcome to the PF!

Convert the vectors into rectangular coordinates, add the x components, add the y components, and convert back to polar form. You will generally always add vectors in rectangular coordinates, so get good at converting back and forth.

Show us what your calculation looks like now.

(And BTW, try to post smaller images in the future so they don't blow up the size of the page. Thanks.)

Yeah sorry about the giant picture :x

Unfortunately I am unsure of exactly how to convert these I really only have experience doing this with a ruler or right triangle in which you just apply SOH-CAH-TOA :/
 
  • #6
berkeman
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http://en.wikipedia.org/wiki/Polar_...rting_between_polar_and_Cartesian_coordinates

i looked at that part and doesn't that only work for right triangles?

Yes. For each vector in polar form (the hypotenuse), there are x and y components that form a right triangle. That's why you use those familiar formulas to convert from polar to rectangular and back again.

Convert each of your two vectors into their rectangular components, and add the vectors component-wise. Then convert back to polar form.

So for the first vector that is straight down, what are its x and y components (you don't even need a calculator for this first vector conversion...)?
 
  • #7
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Yes. For each vector in polar form (the hypotenuse), there are x and y components that form a right triangle. That's why you use those familiar formulas to convert from polar to rectangular and back again.

Convert each of your two vectors into their rectangular components, and add the vectors component-wise. Then convert back to polar form.

So for the first vector that is straight down, what are its x and y components (you don't even need a calculator for this first vector conversion...)?

well the first vector would be (0,35) right? and the second would be (15, 50) I guess. Or I am still not getting this, sorry if I am a little slow I am just totally missing this :/
 
  • #8
berkeman
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9,810
well the first vector would be (0,35) right? and the second would be (15, 50) I guess. Or I am still not getting this, sorry if I am a little slow I am just totally missing this :/

Correct-ish on the first one. Keep in mind that the + x direction is to the right, and the + y direction is up. so the first vector's components would be _____ ?

And two things on the 2nd vector. Draw the x and y components of that vector -- they have nothing to do with the first vector -- just draw the right triangle that has the 2nd vector as the hypotenuse, and show the x and y sides of that triangle. The second thing is that I just noticed that you mislabeled which angle is 25 degrees. Look at your figure, and put the 25 degrees on the correct angle, and then find any other angles that you need in order to use the polar-rectangular conversion formulas.

I have to bain for a while, but will try to check in later. Others will hopefully also chime in if you have further questions.

Also, here's a vector addition tutorial that may help (which I found by googling vector addition tutorial):

http://zonalandeducation.com/mstm/physics/mechanics/vectors/componentAddition/componentAddition2.htm

.
 
  • #9
6
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Correct-ish on the first one. Keep in mind that the + x direction is to the right, and the + y direction is up. so the first vector's components would be _____ ?

And two things on the 2nd vector. Draw the x and y components of that vector -- they have nothing to do with the first vector -- just draw the right triangle that has the 2nd vector as the hypotenuse, and show the x and y sides of that triangle. The second thing is that I just noticed that you mislabeled which angle is 25 degrees. Look at your figure, and put the 25 degrees on the correct angle, and then find any other angles that you need in order to use the polar-rectangular conversion formulas.

I have to bain for a while, but will try to check in later. Others will hopefully also chime in if you have further questions.

Also, here's a vector addition tutorial that may help (which I found by googling vector addition tutorial):

http://zonalandeducation.com/mstm/physics/mechanics/vectors/componentAddition/componentAddition2.htm

.

[PLAIN]http://img704.imageshack.us/img704/1728/35794242.png [Broken]

how was my degrees on the wrong angle?
Unless i can do the 3, 4, 5 triangle rule with that gray triangle but I doubt that is what i'm supposed to do.
I am completely and hopelessly lost I thank you for trying to help and not just doing it all for me but I honestly have no idea what is happening and that tutorial is just talking about right triangles and i don't have a right and whatever I am too frustrated for this now.
 
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  • #10
13
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The object is originally headed on a vector of 270°. It changes course by -25° from 270°. Thus, the angle should be labeled from the 270° axis to the vector of the newest direction.

You should be able to visually see that the angle you labeled in the OP is not 25°, as it is clearly obtuse.
 

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