# vectors/displacement problem

• Sep 2nd 2008, 02:03 PM
viet
vectors/displacement problem
Quote:

A jogger travels a route that has two parts. The first is a displacement A of 2.90 km due south, and the second involves a displacement B that points due east.
Quote:

(1) The resultant displacement A + B has a magnitude of 4.55 km. What is the magnitude of B?
What is the direction of A + B relative to due south?

(2) Suppose that A - B had a magnitude of 4.55 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?
direction of A - B relative to due south
for part1:
$A^2+B^2+C^2 = 0$

B = $\sqrt{A^2+(-C^2)}$

B = $\sqrt{(2.9)^2+(-4.55)^2}$

B = 5.395 km

some how that answer is wrong, can someone explain?
• Sep 2nd 2008, 02:31 PM
o_O
When you draw the vectors, you form a right triangle:

Code:

    |\     | \  2.9 |  \  4.55     |  \     |    \     V----->         B
By the Pythagorean theorem, $2.9^2 + B^2 = 4.55^2$

For part 2, A - B = A + (-B):

Code:

        /|         / |  A - B /  |  A       /  |     /    |     *<----V       B
The triangle doesn't change from part (1) and so the magnitudes should all be the same (but not the direction).

You can see that the direction of A - B is south and to the west by a few degrees. Use trigonometry to find the upper angle to determine the exact direction.
• Sep 2nd 2008, 03:28 PM
viet
i did the Pythagorean theorem wrong. thanks!