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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. I NTERPRET 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. D EVELOP Using Equation 3.1, we see that in two dimensions, a displacement vector 1 r can be written as, in unit vector notation, 1 1 1 1 1 1 ˆ ˆ ˆ ˆ (cos sin ) x y r r i r j r i j       where 2 2 1 1 1 ( ) ( ) x y r r r   and 1 1 1 1 tan ( / ). y x r r 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 1 1 . r v t We shall take a coordinate system with x axis east, y axis north, and origin at the starting point. E VALUATE (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 1 1 10 ˆ ˆ (40 mi/h) h (6.67 mi) 60 r v t j j Similarly, we have 2 ˆ (5.0 mi) , r i and the time spent on this segment is 2 2 2 / (5.0 mi)/(60 mi/h) 0.083 h 5 min. t r v   Finally, the third segment has length 3 3 3 (30 mi/h) 6h/60 3.0 mi. r v t A unit vector in the southwest direction is 1 ˆ ˆ ˆ ˆ cos225 sin 225 ( ) 2 i j i j      Therefore, 3 1 ˆ ˆ ˆ ˆ (3.0 mi) ( ) ( 2.12 mi) ( 2.12 mi) . 2 r i j i j       These displacements and their sum are shown in the figure. The total displacement is 1 2 3 1 1 ˆ ˆ 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 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 ˆ ˆ (2.88 mi) (4.55 mi) ˆ ˆ (8.22 mi/h) (13.0 mi/h) (21/60) h tot r i j v i j t A SSESS 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 2 (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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