Basic Physics Question
Asked by: plato1000-ga
List Price: $10.00
02 Feb 2003 15:07 PST
Expires: 04 Mar 2003 15:07 PST
Question ID: 156463
I have the answer to this problem but can not figure out the math or rationale to make this work: Dan drives 1 mile north. then .50 mile east, 1.0 mile south, 2.0 miles west, then .50 miles northwest and 0.75 miles northeast. What is the magnitude and direction of Dan's displacement vector at the end of the trip? The answer is D vector =1.6 miles and direction is 57 degrees West of North. Please show vector diagram and all calculations used.
Re: Basic Physics Question
Answered By: bio-ga on 02 Feb 2003 18:00 PST
Hi, We can attack the question by first calculating the X and Y components of the movement. I am assuming the directions in the usual way, ie north is positive y, and east is positive x direction. First, please note that the length of X (or Y, for that purpose) component of a 45 degree vector (NW, NE etc) is: |X| = |Y| = length * cos(45) Let's start with adding up the X and Y components: X = 0.5 - 2 - 0.5 * cos(45) + 0.75 * cos(45) = -1.5 + 0.25 * cos(45) = -1.323223 Y = 1 - 1 + 0.5 * cos(45) + 0.75 * cos(45) = 1.25 * cos(45) = +0.883883 So, Dan effectively drove 0.883883 miles to north and 1.323223 miles to west. Total displacement can be calculated using Pythagoras' equation: d^2 = (1.323223)^2 + (0.883883)^2 d = 1.6 miles The angle alpha can be calculated using: tan(alpha) = 1.323223 / 0.883883 alpha = 56.25 degrees Note: You can find a diagram summarizing the vectors at: http://www.geocities.com/bio_ga/156453.html Hope this helps Regards Bio
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