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NOTEBOOK PAGE 16 – 9/7-8/2010

#3. NOTEBOOK PAGE 16 – 9/7-8/2010. Page 16 & 17. Wed 9/8. Tue 9/7. Problem Workbook. 17. 16. Geometry & Trigonometry. P19 #2 P19 # 4 P20 #5 P20 # 7. Write questions!. Write a Summary!. Vector addition & Geometry. Pythagorean theorem

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NOTEBOOK PAGE 16 – 9/7-8/2010

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  1. #3 NOTEBOOK PAGE 16 – 9/7-8/2010

  2. Page 16 & 17 Wed 9/8 Tue 9/7 Problem Workbook. 17 16 Geometry & Trigonometry P19 #2 P19 # 4 P20 #5 P20 #7 Write questions! Write a Summary!

  3. Vector addition & Geometry • Pythagorean theorem • The square of the hypotenuse is equal to the sum of the squares of the other two sides • R2 = x2 + y2 • R = (x2 + y2)1/2

  4. Vector addition & Geometry Given the vectors below Find the magnitude of the Resultant: ON YOUR SLATE! B D A • A + C • A + B • B + H C DRAW Graphical addition on your plastic slate! Show calculations to find magnitude of resultant.

  5. Use the Pythagorean theorem • When you can form a right triangle from the information given. • You know the length of two sides of the triangle. • The vectors are of the same unit of measure.

  6. Vectors and Trigonometry • The Sine value of a right triangle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse of the triangle. • Cosine is a similar ratio • The angle is called Theta (q).

  7. Example: • If hyp = 6 and opp = 4 • Sine q = 4/6 = .66 • This tells us that with the same angle q, the opposite side is always 0.66 times the length of the hypotenuse! • This is very useful!

  8. When do you use Trigonometry? • When a Vector is described by it’s magnitude and direction (angle q) • Since , and • Then by algebra: • Opposite (or y component) = hyp x sin q • Adjacent (or x component) = hyp x cos q • We can always find the x and y components of a vector it’s magnitude qrelative to the x axis! GIVEN: A, then AX = Acosq and AY = Asinq

  9. Resolving a Vector Into Components +y The horizontal, or x-component, of A is found by Ax = A cos q. A Ay q Ax The vertical, or y-component, of A is found by Ay = A sin q. +x By the Pythagorean Theorem, Ax2 + Ay2 = A2. Every vector can be resolved using these formulas, such that A is the magnitude of A, and q is the angle the vector makes with the x-axis. Each component must have the proper “sign” according to the quadrant the vector terminates in.

  10. Vector Components & Trigonometry • Any Vector can be broken down into x and y components. Example: a plane flies @ 100kph at 33O North of West. • V = 100 kph • q = 33o • VX = V cos q • = 100 cos 33 • = 83.9 kph • VY = V sin q • = 100 sin 33 • = 54.5 kph ON YOUR SLATE! GIVEN: A, then AX = Acosq and AY = Asinq

  11. Vector Addition & Geometry Given the vectors below Find the magnitude of the x and y components of each. ON YOUR SLATE! B • 72 km @ 58O NE • 48 km @ 42O SE • 119km @ 26O NE C A Cx = 119cos(26) = 107km Cy = 119sin(26) = 52km Ax = 72cos(58) = 38km Ay = 72sin(58) = 61km Bx = 48cos(42) = 36km By = 48sin(42) = -32km GIVEN: A, then AX = Acosq and AY = Asinq

  12. Analytical Method of Vector Addition 1.Find the x- and y-components of each vector. Ax = A cos q = Ay = A sin q = By = B sin q = Bx = B cos q = Cx = C cos q = Cy = C sin q = Rx= Ry = 2.Sum the x-components. This is the x-component of the resultant. 3.Sum the y-components. This is the y-component of the resultant. 4.Use the Pythagorean Theorem to find the magnitude of the resultant vector. Rx2 + Ry2 = R2

  13. Vector addition & Geometry Given the vectors below Find the magnitude and direction of the sum of the following vectors by using x and y component addition. ON YOUR SLATE! B • 72 km @ 58O NE • 48 km @ 42O SE • 119km @ 26O NE C A Ax = 38km Ay = 61km Q = Tan-1 Ry/Rx = 24O NE By = -32km Bx = 36km Cy = 52km Cx = 107km R= RX = 181km, RY = 81km R= (181km2 + 81km2)1/2 = 198km

  14. 5. Find the reference angle by taking the inverse tangent of the absolute value of the y-component divided by the x-component. q = Tan-1Ry/Rx 6. Use the “signs” of Rx and Ry to determine the quadrant. NW NE (-,+) (+,+) (-,-) (-,+) SW SE

  15. Sample Problem A plane flies 65OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point.

  16. Vector addition & Geometry A plane flies 25OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point. ON YOUR SLATE! C B A • A = 30km @ 25O East of North • B = 30km @ 15O North of East • C = 30km @ 32O South of East • A = 30km @ 65O Northof East _______________! ____!

  17. Vector addition & Geometry A plane flies 25OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point. ON YOUR SLATE! C B AX = 30km cos 65O NE = 12.7km BX = 30km cos 15O NE = CX = 30km cos -32O NE = AY = 30km sin 65O NE = BY = 30km sin 15O NE = CY = 30km sin -32O NE = A

  18. Page 16 & 17 Wed 9/8 Tue 9/7 17 16 Geometry & Trigonometry Practice Problems From Problem Workbook. G.U.E.S.S. P19 #2, 4 P20 #5, 7 Write questions! ! Write a Summary!

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