Question

8 10 11 12

8. In unit-vector notation, A = (4.0 m) + (3.0 m)ŷ and B = (-13.0 m)Â + (7.0 m), (a) what is the sum of the two vectors? (b) What are the magnitude and (c) the direction of A + B (relative to )? 9. Two vectors are given by A = (4.0 m) – (3.0 m)ỹ + (1.0)2 and B = (-1.0 m) + (1.0 m)ĝ+ (4.0 m)2. In unit-vector notation, find (a) A + B, (b) A -B, and (c) a third vector C such that A - B + C = 0. 10. Two vectors are given by A = (4.0 m) + (3.0 mg and B = (-13.0 m)X + (7.0 m) 2. What are the magnitude and direction of A-B. 11. A pilot decides to take his small plane for a Sunday afternoon excursion. He first files north for 155.3 km, then makes a 90° turn to his right and files on a straight line for 62.5 km, then makes another 90° turn to his right and flies 47.5 km on a straight line. (a) How far away from his home airport is he at this point? (b) in which direction does he need to fly from this point on to make it home in a straight line? (c) What was the farthest distance he was away from the home airport during the trip? 12. You are walking in the Lumpinee park heading southwest from your starting point, for 1.72 km. You reach a pool so you make a 90° right turn and walk another 3.12 km to a bridge. How far away are you from you stating point?

Answer 1

W . go.) Ã = (4 û +3 ŷ) m B (B û +7 9, m. B [4 -13) â +(3+1) ĝ - -qâ +lo ý m. 1A+B) - S60)*7 (10)? - 5181 = 13454 m. Gids direction ou tam (105) 90+ 48° = 138" then a) As Yü -sý + Ź B -loget loģ +4.0 2 Ā+ B', (4-10 û + 1-3 tilý + (Hot 4.) 2 = ( 3 â -2ỹ +52). A- R. (4+1) +(--)3 + ( 1. -)2 = ( % - - 32) -

No АжE +- 0 ) с. - (А. Е) . - С.ox + 4 + 301 -) Зака од 8.

(2.5 KMB A 47.5 km 155.3 Кн 11.) 1 a.) distance from home auport Tocl = 162.5 + (155.3-47.5)|| - J (62.5)4+ (107.813 = 124.68 Km. b) duiection ou tan" (1078) - 59.891 CI forthest distance - 10B1 - 1 162.1)?+ (1553)? - 167.405 km 12.3 2 distance from starting point - JOBI - JB:12)+(1.72)2 23.56 km (Pythagoras theorum in A OAB) pe 40 32 & 1.72