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magnitude of net displacement?
A jogger runs 10.0 blocks due east, 5.0 blocks due south, and another 2.0 blocks due east. Assume all blocks are of equal size. What is the magnitude of the jogger's net displacement?
How the heck do i do this!? Thank you to all who help out!
Okay, this is a physics question, but I actually know how to help on this one! LOL
First off, draw it out on paper. Draw an arrow to the right, meaning due east, and put 10.0 above it. then from the end of that arrow, draw another arrow straight down (due south), and put 5.0 beside it. Then draw another arrow from the end of the one you just drew to the right (due east) and put 2.0 above it.
Now net displacement is how far from the starting place the jogger is now. The two arrows pointing right (east) can be summed into one arrow pointing right (east). If you add them together, you get one arrow pointing east at 12.0. See?
Now redraw it so that there are only two arrows, one pointing right (put 12.0 above it) and one down (label it with 5.0). Be sure that the one pointing down is connected to the one pointing right. If you were to draw a line connecting the very tail-end of the right arrow to the head of the down arrow, you will have a triangle. This side of the triangle that you can draw is the net displacement. Now just use the pythagorean theorem to find its magnitude. a^2 + b^2 = c^2. Let a = 12.0 and b = 5.0. c = square root of (12.0^2 + 5.0^2). That's it.
I hope this helps.
Joggers Park