Mr. Rockensies – Regents Physics AIM – How do we add vectors? DO NOW – Where have you heard the word vector aside from Physics class? HW - Textbook p. 26 #67(a-d), 68(a-d), 72(a-d) Vector Addition Mr. Rockensies – Regents Physics

Vectors Vectors Scalars (no direction) Quantities with magnitude (amount, size) and direction. Example: 20 m North or 20 m West Vectors Displacement Velocity Acceleration Force Momentum Impulse Electric Field Scalars (no direction) Distance Speed Time Mass Energy/Work Charge Power

NEVER FORGET TO DRAW THE ARROWS!! Drawing Vectors Vectors can be drawn to graphically represent magnitude as well as direction. Length indicates magnitude, and therefore must be drawn to scale using a ruler and protractor. The angle indicates direction, represented by θ (theta). length θ Horizontal Axis = +X direction NEVER FORGET TO DRAW THE ARROWS!!

Resultant – Adding Vectors Resultant – the result of 2 or more displacements (vectors) 20 m West R = resultant displacement θ = direction 20 m North R θ R = 28 m, 45° determined by measuring with a ruler and protractor

Mathematical Techniques When vectors are at right angles, we can use the Pythagorean Theorem and SOHCAHTOA: a2 + b2 = c2 20 m West R2 = (20m)2 + (20m)2 R = √800 m2 R = 28.2 m 20 m North R θ tan θ = opp/adj = 20/20 = 1 θ = tan-1 (1) = 45°

Vector Addition (cont.) Same Direction: simply add = 11 m 4 m 7 m Opposite Direction: subtract 5 m = 4 m 9 m

Practice A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement? A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement?

Practice Problem #1 A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement? **not drawn to scale** 1500 miles θ 200 miles Resultant - 1513 miles a2 + b2 = c2 (1500 m)2 + (200 m)2 = R2 R = √ (1500 m)2 + (200 m)2 R = 1513.275 m tan θ = opp/adj θ = tan-1 (200/1500) θ = 7.5946433°

Practice Problem #2 A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement? By subtracting the opposing directions from each other, we find the hiker’s displacement along the y-axis to be 30 m North, and the displacement on the x-axis to be 10 m West. a2 + b2 = c2 302 + 102 = R2 R = √900 + 100 R = 31.623 m tan θ = opp/adj θ = tan-1 (10/30) θ = 18.435°

Mr. Rockensies – Regents Physics AIM – What are the components of the resultant? DO NOW – A car drives 4 miles North, 3 miles East, and 2 miles South, what is its total displacement? HW - Textbook p. 53 #50, 51, 53 Vector Addition Mr. Rockensies – Regents Physics

Velocity Vectors Occur at the same time – concurrent Displacement vectors occurred sequentially – one after the other Boat velocity Boat How do we find the resultant velocity? Stream velocity River

Resultant velocity found by drawing the vectors head to tail – just as with displacement 8 m/s Boat Boat 6 m/s 8 m/s VR2 = 82 + 62 = 100 VR = 10 m/s tan θ = 6/8 θ = tan-1 (6/8) = 37° θ 6 m/s VR Velocity Resultant

Vector Components If R = A + B, then we can say that A and B are components of R B R Two or more components add to make a resultant A Rectangular Components – components which lie on the x and y axes A resultant can also be resolved back into components!!

Japanese Vector Video Japanese Vector Video - Launching a Ball from a moving truck

Practice Problems A ball is kicked 30° above the horizontal at a velocity of 65 m/s. (a) What is the horizontal component of the velocity? (b) What is the vertical component of the velocity? A frisbee is tossed 12° above the horizontal at a velocity of 23 m/s. (a) What is the horizontal component of the velocity? (b) What is the vertical component of the velocity?

Practice Problem #1 Horizontal Component: Vertical Component Horizontal Component: Since we have the hypotenuse and the angle, we need to find this piece by using the cosine trig function Ax=Acos(θ) Ax=(65m/s)cos(30°) Ax=56.29 m/s (b) Vertical Component: Since we have the hypotenuse and the angle, we need to find this piece by using the sine trig function Ay=Acos(θ) Ay=(65m/s)sin(30°) Ay=32.5 m/s 65 m/s 30° Horizontal Component

Practice Problem #2 Horizontal Component: Since we have the hypotenuse and the angle, we need to find this piece by using the cosine trig function Ax=Acos(θ) Ax=(23m/s)cos(12°) Ax=22.50 m/s (b) Vertical Component: Since we have the hypotenuse and the angle, we need to find this piece by using the sine trig function Ay=Acos(θ) Ay=(23m/s)sin(12°) Ay=4.78 m/s 23 m/s 12°