Posted by Autumn on Monday, September 7, 2009 at 8:46pm.
The best way to solve this type of problems is by a table of component vectors.
Let x- and y-axes represent east and north respectively.
Resolve each segment of the trip into the x- and y-components.
Add up vectorially the components, which is the resultant displacement.
If the displacement is (x,y), the magnitude is √(x²+y²).
Work in hours, i.e. minutes should be divide by 60.
Distance x-component y-component
59.78 59.78*cos(0) 59.78*sin(0)
89.67 89.67*cos(90) 89.67*sin(90)
29.89 29.89*cos(270-59.78) 29.89*sin(270-59.78)
--------------------------------------
sum x y
Find (x,y), and the magnitude of displacment is √(x²+y²)
Divide the magnitude of displacement by the time (in hours) will give you the average velocity in miles per hour.
Post your answer for verification if necessary.
i got 16.46 mi/hr. is that right??
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