VECTORS!!!.

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VECTORS!!!

Trig Practice 1. 2. y 100 m. x 40 O = ? Fx = ____ F = 10 lbs. A = ? 60 100 m. O = ? A = ? 2. y x F = 10 lbs. 40 Fx = ____ Fy = ____ 3. 60 H = ____ O = ____ A = 22

Trig Practice - ANSWERS 1. 60 100 m. O = ? A = ? 2. y x F = 10 lbs. 40 Fx = ____ Fy = ____ 0 = 100sin(60) = 86.6 A = 100cos(60) = 50 3. 60 H = ____ O = ____ A = 22 H = 38/(sin(60)) H = 44 O = 22 tan(60) O = 38 Fx = -10cos(40)= -7.66 Fy = -10sin(40) = -6.43 Fx and Fy ARE CALLED VECTOR COMPONENTS

Trig Practice 4. Earth in March 1 q = degree To a Distant Star 3600 Sun Earth in March Earth in December To a Distant Star q = degree 1 3600 Find the distance to the star if the Sun-Earth distance is 1.49 X 1011 meters.

Trig Practice - ANSWER EARTH sin(1/3600) = (1.49 x 1011 / D) 4. Sun Earth in March Earth in December To a Distant Star q = degree 1 3600 Find the distance to the star if the Sun-Earth distance is 1.49 X 1011 meters. EARTH sin(1/3600) = (1.49 x 1011 / D) D = 3.1 x 1016 m 1.49 X 1011 m Θ NOTE: Θ is so small that both other angles are ~90º SUN

Vector Addition A mail carrier drives 22km to the next town, due N. Then, drives 60 º South of East for 47km. What is the displacement from the post office? HINT: Divide each vector, A and B, into vector components, Ax and Ay, and Bx and By. A B Resultant Vector = R

Vector Addition A mail carrier drives 22km to the next town, due N. Then, drives 60 º South of East for 47km. What is the displacement from the post office? Ax = 0 Ay = 22km Bx = 47cos60 = 23.5 By = 47sin60 = -40.7 Rx = 0 + 23.5 = 23.5 Ry = 22 – 40.7 = -18.7 R = √(Rx2 + Ry2) = 30km tanΘ = Ry/Rx = Θ=-38.5º A B Resultant Vector = R

Vector Addition Travel 12m 37º East of North. Travel 15m 40º South of East. Travel 6m 60º South of West. Draw the 3 vectors and their components. Determine the final location of the traveler: Both displacement from the start AND angle.

Vector Addition Travel 12m 37º East of North. Travel 15m 40º South of East. Travel 6m 60º South of West. A C B

Vector Addition Travel 12m 37º East of North. Travel 15m 40º South of East. Travel 6m 60º South of West. Ax = 12sin37 = 7.2 Ay = 12cos37 = 9.6 Bx = 15cos40 = 11.5 By = 15sin40 = 9.6 in the –y direction Cx = 6cos60 = 3 in the –x direction Cy = 6sin60 = 5.2 in the –y direction X = Ax + Bx + Cx = 15.7 Y = Ay + By + Cy = -5.2 A C B

Vector Addition Travel 12m 37º East of North. Travel 15m 40º South of East. Travel 6m 60º South of West. X = Ax + Bx + Cx = 15.7 Y = Ay + By + Cy = -5.2 Final = (X2 + Y2)½ = 16.5m A 15.7 5.2 C B tanΘ = 5.2/15.7 Θ = 18.3º South of East

#1 If you drive west at 20 km/h for one hour, then drive east at 15 km/h for one hour, your net displacement will be: 5 km east. 35 km west. 35 km east. 5 km west.

#1 If you drive west at 20 km/h for one hour, then drive east at 15 km/h for one hour, your net displacement will be: 5 km east. 35 km west. 35 km east. 5 km west.

x (positive), y (positive). x (positive), y (negative). #2 A 200-lb force is pulling on an object, as shown. The sign of the x and y components of the force are: x (positive), y (positive). x (positive), y (negative). x (negative), y (positive). x (negative), y (negative).

x (positive), y (positive). x (positive), y (negative). #2 A 200-lb force is pulling on an object, as shown. The sign of the x and y components of the force are: x (positive), y (positive). x (positive), y (negative). x (negative), y (positive). x (negative), y (negative).

#3 Three boys each pull with a 20-N force on the same object. The resultant force will be zero. 20 N to the left. 20 N up. 20 N down.

#3 Three boys each pull with a 20-N force on the same object. The resultant force will be zero. 20 N to the left. 20 N up. 20 N down.

Vg will always equal Va + V Vg can be greater than Va + V #4 Consider a plane flying with groundspeed Vg and airspeed Va in a wind with speed V. Which of the following relationships is true? Vg will always equal Va + V Vg can be greater than Va + V Vg can be less than Va – V Vg can have any value between Va + V and Va - V

#4 Consider a plane flying with groundspeed Vg and airspeed Va in a wind with speed V. Which of the following relationships is true? Vg will always equal Va + V Vg can be greater than Va + V Vg can be less than Va – V Vg can have any value between Va + V and Va – V The direction of the wind was not given to the airspeed may be helped or hindered by the wind speed.

The direction is changing. #5 If the acceleration vector of a body is perpendicular to the velocity vector, which of the following must be true? The speed is changing. The direction is changing. Both the speed and the direction are changing. The direction is not changing.

The direction is changing. #5 If the acceleration vector of a body is perpendicular to the velocity vector, which of the following must be true? The speed is changing. The direction is changing. Both the speed and the direction are changing. The direction is not changing. a v

equal to the magnitude of either one. #6 If the sum of two vectors equals zero, the magnitude of their difference is: equal to the magnitude of either one. equal to twice the magnitude of either one. less than twice the magnitude of either one.

equal to the magnitude of either one. #6 If the sum of two vectors equals zero, the magnitude of their difference is: equal to the magnitude of either one. equal to twice the magnitude of either one. less than twice the magnitude of either one. DIFFERENCE SUM DIFFERENCE MEANS TO REVERSE DIRECTION OF 2ND VECTOR