Slide 1 - Vector Solutions

• With no right angles

Slide 2 - A plane cruises a 425 km/hr on a heading of 50.0o, while encountering a wind blowing towards 325o at 135 km/hr. What is the plane’s velocity over ground?

Slide 3 - A plane cruises a 425 km/hr on a heading of 50.0o,

• while encountering a wind blowing towards 325o at 135 km/hr.
• What is the plane’s velocity over ground?

Slide 4 - Component Method

• Break each vector into X and Y components
• Add the X’s and add the Y’s
• Combine the Total X and Total Y into a Resultant

Slide 5 - A plane cruises a 425 km/hr on a heading of 50.0o,

• 50o
• Y
• X
• 425 km/hr

Slide 6 - A plane cruises a 425 km/hr on a heading of 50.0o,

• 50o
• Y = (425km/hr)(cos50o)
• = 273 km/hr
• X = (425km/hr)(sin50o)
• = 326 km/hr
• 425 km/hr
• X
• Y
• Y
• X

Slide 7 - while encountering a wind blowing towards 325o at 135 km/hr.

Slide 8 - while encountering a wind blowing towards 325o at 135 km/hr.

• Y
• X
• 135 km/hr
• 35o

Slide 9 - while encountering a wind blowing towards 325o at 135 km/hr.

• Y = (135km/hr)(cos35o)
• = 111 km/hr
• X= (135km/hr)(sin35o)
• = 77.4 km/hr
• 135 km/hr
• 35o

Slide 10 - Combine the Y-components

• Y = 111 km/hr
• Y = 273 km/hr
• Ytotal = 384 km/hr

Slide 11 - X= -77.4 km/hr

• Combine the X-components
• X = 326 km/hr
• Xtotal = 249 km/hr

Slide 12 - Find the Resultant

• Xtotal = 249 km/hr
• Ytotal = 384 km/hr

Slide 13 - Find the Resultant

• C2 = A2 + B2
• C =  (384 km/hr)2 + (249 km/hr )2
• C= 458 km/hr
• 
• = tan-1 opp
• adj
• = tan-1 249 km/hr 384 km/hr
• = 33.0o
• A = 384 km/hr
• B = 249 km/hr
• B
• A
• C
• Answer: 458 km/hr @ 33.0o

Slide 14 - Use the Law of Cosines and the Law of Sines to solve the parallelogram

Slide 15 - A plane cruises a 425 km/hr on a heading of 50.0o,

• while encountering a wind blowing towards 325o at 135 km/hr.
• What is the plane’s velocity over ground?
• 50.0o
• 325o
• 85o
• 425 km/hr
• 135 km/hr

Slide 16 - Complete the parallelogram

• 85o
• 85o
• 95o
• 95o

Slide 17 - Law of Cosines

• 95o
• b
• a
• c
• c =  a2 + b2 - 2abcosC

Slide 18

• 95o
• b = 425 km/hr
• a = 135 km/hr
• c
• c =  a2 + b2 - 2ab cosC
• c =  (135 km/hr)2 + (425 km/hr)2 - 2(135 km/hr)(425 km/hr) cos(95o)
• c = 457 km/hr

Slide 19 - Law of Sines

• sin B = sin C
• b c
• B = sin-1 (425 km/hr) sin (95o)
• (457 km/hr)
• B = 67.9o
• 95o
• b = 425 km/hr
• a = 135 km/hr
• B
• C = 457 km/hr

Slide 20 - c

• Direction of Resultant = 325o + 67.9o = 393o
• Since North is 360o, the Resultant is pointing at 33o
• B
• 325o
• 33o

Slide 21 - 457 km/hr

• A plane cruises a 425 km/hr on a heading of 50.0o,
• while encountering a wind blowing towards 325o at 135 km/hr.
• What is the plane’s velocity over ground?
• Answer: 457 km/hr @ 33o
• 33o