Algebraic Vectors A) an (X,

Algebraic Vectors

a) an (x, y) coordinate system would be appropriate; movement will occur both along the sidewalk (i.e. down the street) and across the sidewalk (between the buildings/yards and the curb), requiring a two-dimensional coordinate system

b) a simple x coordinate system or number line is sufficient; the tightrope walker will only be moving in a single direction (albeit wit positive and negative relative magnitude)

c) as stated, only an (x, y) coordinate system would be needed, but in reality a three-dimensional (x, y, z) coordinate system is required to plot submarine and aircraft movements, as travel can occur along the two dimensions of the cardinal directions (i.e. forward-back or North-South and right-left or East-West), as well as vertically (up and down).

b.

3.

a.

b.

Without resolving vectors into components, no understanding of relative motion can actually be established. Vectors are inconsequential if not stated and understood in relation to a fixed point (usually the starting point of the vector).

35m

15m

=25(

=65(

a b

15/sin90 = a/sin25 15/sin90 = b/sin65

15 x sin25 = a 15 x sin65 = b

6.34=a 13.59 = b d+b = 48.59 a = 6.34

48.592+6.342 = 2401.18

( 2401.18 = 49

c = 49 = magnitude

49/sin90 = 6.34/sinA

49 x sinA = 6.34

sinA = 6.34/49

sinA = .129

A ( 7.43( = direction d

c

A

2.5km

5.2km

A

B

B = 180 -- (35-22)

B = 173(

sinB = .122

b2 = 2.52 + 5.22 -- (2x 2.5 x 5.2 x cos173)

b2 = 58.71

b = 7.66km = magnitude

7.66/sin173 = 5.2/sinA

62.85 = 5.2/sinA

sinA = 5.2/62.85

sinA = .083

A = 4.75

35 -- a = 30.25 = direction

Displacement is 7.66km at 30.25 degrees above the horizon

C

b

Displacement is 49m at 7.43 degrees East of North (assuming North for downfield)

8m

3.5m

5m

C

A

B

a c b

D

C = 90 E = 180 -- D = 35

D = 90 + 55 = 145 F = 180 -- E -- C = 55

3.5/sin90 = f/sin55 3.5/sin90 = e/sin35

3.5 x sin55 = f 3.5 x sin35 = e f = 2.87 e = 2.01

b = f + 8 = 10.87 a = e + 5 = 7.01

c2 = 10.872 + 7.012

c2 = 167.297

c = 12.93 = magnitude

12.93/sin90 = 7.01/sinA

12.93 x sinA = 7.01

sinA = 7.01/12.93 = .54

A = 32.8 = direction

Displacement is 12.93 meters 32.8 degrees East of North

E

F

f e

60 km

c

C

A

B

D = 30 E = 60

F = 90 -- D = 60 C = 180 -- E -- G

G = 180 -- 90 -- F = 30 C = 180 -- 60 -- 30 = 90

c2 = 602 + 1552

c2 = 27625

c = 166.201 = magnitude

166.201/sin90 = 155/sinA

166.201 x sinA = 155

sinA = 155/166.201 = .933

A = 68.84 = direction

Displacement is 166.201 km at 68.84 degrees West of North

D

E

F

G

5m/s

3m/s c2 = 32 + 52 5.83/sin90 = 3/sinA

c2 = 9 + 25 = 34 5.83 x sinA = 3

c = 5.83 = magnitude sinA = 3/5.83

sinA = .514

A = 30.97 = direction

Displacement is 5.83 meters/second at 30.97 degrees downstream of perpendicularity with the flow of the river

A

c

1m/s

6m/s c