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Unformatted text preview: Solved Problems for chapter 3 9. Picture the Problem : The base runner travels from C (home plate) to first base, then to A (second base), then to B (third base), and finally back to C (home plate). Strategy: The displacement vector ∆ r r is the same as the position vector r r if we take home plate to be the origin of our coordinate system (as it is drawn). The displacement vector for a runner who has just hit a double is drawn. Solution: 1. (a) Write the displacement vector from C to A in terms of its x and y components: ( 29 ( 29 ˆ ˆ 90 ft 90 ft = + r x y r 2. (b) Write the displacement vector from C to B in terms of its x and y components: ( 29 ( 29 ( 29 ˆ ˆ ˆ 0 ft 90 ft 90 ft = + = r x y y r 3. (c) For a home run the displacement is zero: ( 29 ( 29 ˆ ˆ 0 ft 0 ft = + r x y r Insight: The displacement is always zero when the object (or person) returns to its original position. 21. Picture the Problem : The vectors involved in the problem are depicted at right. The control tower (CT) is at the origin and north is up in the diagram. Strategy: Subtract vector B r from A r using the vector component method. Solution: 1. (a) A sketch of the vectors and their difference is shown at right. 2. (b) Subtract the x components: ( 29 ( 29 ( 29 ( 29 220 km cos 180 32 140 km cos 90 65 310 km x x x D A B = = °  ° =  3. Subtract the y components: ( 29 ( 29 ( 29 ( 29 220 km sin 180 32 140 km sin 90 65 57 km y y y D A B = = °  ° = 4. Find the magnitude of D : ( 29 ( 29 2 2 2 2 5 310 km 57 km 320 km 3.2 10 m x y D D D = + = + = = × 5. Find the direction of D : 1 1 57 km tan tan 10 180 170 or 10 north of west...
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This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.
 Spring '11
 BEAN
 pH

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