Name: Date: AP Physics 1 Per. Vector Addition Practice. 1. F1 and F2 are vectors shown below (N is a unit of force, it stands for Newton, not north)

Transcription

1 ame: Date: AP Phsics 1 Per. Vectr Additin Practice 1. F1 and F are ectrs shwn belw ( is a unit f frce, it stands fr ewtn, nt nrth) F1 = 500 F = a) Add the ectrs F1 and F: F1+F = 1. Add graphicall and numericall umerical Additin (b cmpnents) F F = Graphical Additin (tail t tip) 1 = tan = abe ais 551 If u measure the length f 1 n the graph 1 F1 = F = 300 scale and measure with a prtractr, u will get the same result as abe: = abe ais 50

2 b) ubtract the ectr F frm F1: F1 F = umerical ubtractin (b cmpnents) = 9 abe ais Graphical ubtractin (F1 and F tail t tail O F1 and F tail t tip) F1 = 500 F F F = D1 = 0.4km -F F1. Ale walks 0.40 km in a directin 60 west f nrth, then ges 0.50 km due west. Find the displacement b adding the ectrs mathematicall. D =91 = tan D =0. =-0.85 D = 0.5km 0.87km tan = 0.87 km, 13 f 0.

3 3. A hiker s trip cnsists f three segments. Path A is 8.0 km lng heading 60 nrth f east. Path B is 7.0 km lng in a directin 30 nrth f west. Path C is 4.0 km lng heading 70 east f suth. Find the displacement f the hiker. = 8km C = 1.7 =9.06 = 7km = 85 km, 79.4 f 4. A car dries 60.0 miles directl nrth in ne hur and then turns and dries 80.0 miles directl west in ne hur and 15 minutes. D = 80mi a) hat is the ttal distance the car dre? 140 mi 30 85km tan ,7 60 C = 4km 9.06 b) hat is the car s ttal displacement fr the trip? (magnitude and directin) (pthagream s Thm) D = 100 mi, 53 f c) hat was the car s aerage elcit fr the trip? D 100mi a 44.4mph, 53 t.5hr (magnitude and directin) d) hat was the car s aerage speed fr trip? f D s d 140mi 6. mph t.5hr D1 = 60mi

4 5. Add tgether the fllwing ectrs graphicall and numericall (b cmpnent), giing the magnitude and directin f the resultant and the equilibrant. Vectr A: (frm + ais) Graphical methd: Vectr B: (frm + ais) Vectr C: (frm + ais) C Cmpnent methd: B B =Bsin10 10 C =Ccs60 60 C =Csin60 C B =Bcs10 A 60 A =Asin60 A =Acs60 A B C A B 599.1m tan A car dries 50. m in a directin 35.0 uth f east. a) Hw far uth did the drie? cmpnent f displacement suth is 143 m (50sin35) b) Hw far east did the drie? cmpnent f displacement east is 05 m (50cs35) 35

5 The car then turns and dries fr 400. m in a directin 65.0 nrth f east. c) hat is the ttal displacement in the nrth/suth directin fr the tw parts f the trip? D D1 D m d) hat is the ttal displacement in the east/west directin fr the tw parts f the trip? D D1 D m e) hat is the magnitude and directin f the ttal displacement f the car fr the entire trip? D D = 0m 434m D = 374m tan = 434 km, 30 f 7. A plane aims nrth and mes with a elcit f 00 m/s (relatie t the air) (PA). The wind blws t the east at 30 m/s (relatie t the grund) (AG). The plane s elcit relatie t the grund (PG) is equal t the sum f these tw elcities (VPG = PA + AG). Calculate the plane s ttal elcit. PG AG PA PG PG tan PA AG 0m / s 81.5 PA AG f

6 8. A rier flws nrth at 4.00 m/s (relatie t the grund) (G). A bat aims east with a elcit f 6.00 m/s (relatie t the water) (B) tring t crss the rier. a) hat is the bat s ttal elcit (BG)? B = 6m/s G = 4 m/s b) if the rier is 100. m wide, hw lng des it take the bat t get acrss the rier? The elcit f the bat relatie t the grund has perpendicular cmpnents and these are independent f each ther. The bat mes directl acrss the rier (east) because f its mtr and bcause it is aimed east. Bat simultaneusl mes dwnstream because f the current. These tw perpendicular parts f the bats mtin are independent f each ther. An alteratin in ne f the cmpnents will nt affect the ther cmpnent. Fr instance, if the water current increased, then the bat wuld still be cering grund in the easterl directin at the same rate. It is true that the increase f BG = 6m/s the water elcit wuld cause the bat t trael mre nrthward; hweer, the bat still traels eastward at the same speed. Perpendicular cmpnents f mtin d nt affect each D t t c) hw far dwnstream des the bat end up D BG BG t s 6 BG PG tan B 7.m / s 33.7 G B f G BG = 4 m/s D t BGt 4(16.7) 66. 8m