A2 2010 PHYS115-handoutsolutions

A2 2010 PHYS115-handoutsolutions - F10-PHYS115-UAE 27....

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F10-PHYS115-UAE Assignment 2 Solutions 1/4 27. INTERPRET We interpret this as a problem involving the addition of three displacements in two dimensions and finding the magnitude and direction of the resultant vector. The key concepts here are displacement and average velocity. DEVELOP Using Equation 3.1, we see that in two dimensions, a displacement vector 1 r can be written as, in unit vector notation, 11 1 1 1 1 ˆˆ ˆ ˆ (cos sin ) xy rr i r jr i j     where 22 1 () r    and 1 1 tan ( / ). yx  One may write down a similar expression for 2 r and 3 . r The displacement vector 1 r is related to the velocity vector 1 v by . rv t  We shall take a coordinate system with x axis east, y axis north, and origin at the starting point. EVALUATE (a) The first segment of the trip which last, for 1 10 min (1/6) h, t  can be written, in unit-vector notation, as 1 10 (40 mi/h) h (6.67 mi) 60 t j j      Similarly, we have 2 ˆ (5.0 mi) , ri  and the time spent on this segment is 2 / (5.0 mi)/(60 mi/h) 0.083 h 5 min. tr v Finally, the third segment has length  33 3 (30 mi/h) 6h/60 3.0 mi. t A unit vector in the southwest direction is 1 ˆ ˆ cos 225 sin 225 ( ) 2 ij i j   Therefore, 3 1 ˆ ˆ (3.0 mi) ( ) ( 2.12 mi) ( 2.12 mi) . 2 j i j     These displacements and their sum are shown in the figure. The total displacement is 123 1 1 ˆ ˆ ˆ ˆ (6.67 mi) (5.0 mi) ( 2.12 mi) ( 2.12 mi) (2.88 mi) (4.55 mi) tot r r v t j i i j i j  (b) Since the total travel time is 10 min 5 min 6min 21 min (21/60) h, t the average velocity for the trip is (2.88 mi) (4.55 mi) (8.22 mi/h) (13.0 mi/h) (21/60) h tot r vi j t ASSESS We expect both tot r and v to be in the first quadrant since their components are all positive. Instead of unit-vector notation, tot r and v could be specified by their magnitudes (2.88 mi) (4.55 mi) 5.38 mi tot r and 15.4 mi/h, v respectively, and common direction, 1 tan [(4.55 mi)/(2.88 mi)] 57.7 N of E.
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F10-PHYS115-UAE Assignment 2 Solutions 2/4 33. INTERPRET This is a problem involving relative velocities. The quantity of interest is the wind velocity relative to the ground.
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This note was uploaded on 10/05/2010 for the course PHYS 115 taught by Professor Rohan during the Fall '06 term at Waterloo.

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A2 2010 PHYS115-handoutsolutions - F10-PHYS115-UAE 27....

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