NationStates Jolt Archive


acceleration vectors

Shqipes
29-01-2006, 21:54
last question i promise

http://i26.photobucket.com/albums/c144/slicknick727/vectors2.jpg

the question:
http://i26.photobucket.com/albums/c144/slicknick727/question.jpg

i tried to find the x component by doing cos(15)=ax/2 and got 1.93, but this is the wrong answer
Tactical Grace
29-01-2006, 22:06
Interesting one. Think of it as an entertaining puzzle, not a chore. Learn to love the challenge. You will find the hours fly by. :)
The Infinite Dunes
29-01-2006, 22:06
wouldn't you be using cos or sin or whatever of 30 and 60? 45-15= 30 and 90-30=60

... making the answers root 3 and 1.
Shqipes
29-01-2006, 22:09
wouldn't you be using cos or sin or whatever of 30 and 60? 45-15= 30 and 90-30=60

im sorry i dont really understand. maybe the problem is the fact that i dont know exactly whree to place the a(parallel) and a(perpendicular) vectors
Shqipes
29-01-2006, 22:11
is this how theyre placed?

http://i26.photobucket.com/albums/c144/slicknick727/abectors.jpg
The Infinite Dunes
29-01-2006, 22:12
Well... think about it... I've given you the answers. Try and figure out how i got them. I've even told you how to get the angle you need to solve the problem. If I tell you anymore then you won't learn anything.
Tactical Grace
29-01-2006, 22:14
Draw a line from the centre of the circle, through the car, and out of the side.

That's one component.

Now draw a line at 90 degrees to that one, through the car lengthways.

That's another component.

EDIT: I think you need to revisit your terminology here, and judging by your difficulties with the constant acceleration equations, go through the problem-solving method with your teacher.
Shqipes
29-01-2006, 22:16
Well... think about it... I've given you the answers. Try and figure out how i got them. I've even told you how to get the angle you need to solve the problem. If I tell you anymore then you won't learn anything.

thanks man, ill see how you got those
The Infinite Dunes
29-01-2006, 22:16
is this how theyre placed?No. The tangent is the line in which the car is pointing towards. It's perpendicular is 90 degrees to that. For heavens sake those lines aren't even perpendicular.
Tactical Grace
29-01-2006, 22:18
No. The tangent is the line in which the car is pointing towards. It's perpendicular is 90 degrees to that. For heavens sake those lines aren't even perpendicular.
That's what I meant about terminology. There is really no hope of solving anything if people don't know what the tools are called.
Shqipes
29-01-2006, 22:18
o ok, ill try again
Tactical Grace
29-01-2006, 22:21
like this?
:rolleyes:

One line is drawn for you. The 45 degree line. The one at 45 degrees to the horizontal. Extend it through the car out of the circle. There's your perpendicular. The tangent is at 90 degrees, lying along the car's long axis.
Shqipes
29-01-2006, 22:22
hmmm so it would be like this right

http://i26.photobucket.com/albums/c144/slicknick727/vectors2a.jpg
Tactical Grace
29-01-2006, 22:26
hmmm so it would be like this right
That's more like it.
Shqipes
29-01-2006, 22:28
now, i know a perpendicular would be 1.73 because (cos(30))(2)=1.73, but would it be a negative or positive number since it is ponting downwards
Shqipes
29-01-2006, 22:29
That's more like it.

thanks
The Infinite Dunes
29-01-2006, 23:08
now, i know a perpendicular would be 1.73 because (cos(30))(2)=1.73, but would it be a negative or positive number since it is ponting downwards
You have no origin to relate to... so the're magnitudes really...