A person walks the path shown below. The total trip consists of four straight-line paths.
Path 1 is 90.0 m due east.
Path 2 is 160.0 m due south.
Path 3 is 120.0 m 30.0° south of west.
Path 4 is 200.0 m 60.0° north of west.
At the end of the walk, what is the person's resultant displacement?
___ m at ___° south of west.
2006-09-27 18:11:58 · 4 answers · asked by tingerpoo 2 in Science & Mathematics ➔ Mathematics
I have plotted your path and have shown the calculation here:
http://img154.imageshack.us/img154/2796/plotjw2.png
Please check that I have followed the path instructions and verify for yourself the lengths of the segments I have indicated.
2006-09-27 19:24:06 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
Take a set of coordinate axes with x axis giving the east-west direction and y axis giving north-south direction.
Positive x axis gives east direction, positive y axis gives north direction.
Now, draw the vectors, each of them with their tail at the origin.
Next, resolve the vectors into x and y components.
Next, add all the vectors by adding the x components together and the y components together.
The sum is the net displacement.
90 m due east = 90 i + 0 j
160 m due south = 0 i - 160 j
120 m 30* SW = -120cos30 i - 120sin30 j
= -60sqrt(3) i - 60 j
200 m 60* NW = -200cos60 i + 200sin60 j
= -100 i + 100sqrt(3) j
Net displacement = (90 - 60sqrt(3) - 100) i + (-160 -60 +100sqrt(3) ) j
Square the components, add and take the square root.to find out the magnitude.
2006-09-28 01:28:07 · answer #2 · answered by astrokid 4 · 0⤊ 0⤋
distance covered =123.1546 meters from origin at the direction of 22.32 degrees south of west.
please give me best answer. i need it.
2006-09-28 01:36:46 · answer #3 · answered by ravi 1 · 0⤊ 0⤋
How about you try this one on your own since we just answered a very similar question for you?
2006-09-28 01:20:43 · answer #4 · answered by Anonymous · 0⤊ 0⤋