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## VECTORS

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**VECTORS**Honors Section 1.5 – 1.9 Regular Physics Chapter 2**Objectives and Essential Questions**• Objectives • Distinguish between basic trigonometric functions (SOH CAH TOA) • Distinguish between vector and scalar quantities • Add vectors using graphical and analytical methods • Essential Questions • What is a vector quantity? What is a scalar quantity? Give examples of each.**SCALAR**A SCALAR quantity is any quantity in physics that has MAGNITUDE ONLY Number value with units**VECTOR**A VECTOR quantity is any quantity in physics that has BOTHMAGNITUDE and DIRECTION An arrow above the symbol illustrates a vector quantity. It indicates MAGNITUDE and DIRECTION**VECTOR APPLICATION**ADDITION: When two (2) vectors point in the SAME direction, simply add them together. EXAMPLE: A man walks 46.5 m east, then another 20 m east. Calculate his displacement relative to where he started. + 20 m, E 46.5 m, E MAGNITUDE relates to the size of the arrow and DIRECTION relates to the way the arrow is drawn 66.5 m, E**VECTOR APPLICATION**SUBTRACTION: When two (2) vectors point in the OPPOSITE direction, simply subtract them. EXAMPLE: A man walks 46.5 m east, then another 20 m west. Calculate his displacement relative to where he started. 46.5 m, E - 20 m, W 26.5 m, E**NON-COLLINEAR VECTORS**When two (2) vectors are PERPENDICULAR to each other, you must use the PYTHAGOREAN THEOREM FINISH Example: A man travels 120 km east then 160 km north. Calculate his resultant displacement. the hypotenuse is called the RESULTANT 160 km, N VERTICAL COMPONENT START 120 km, E HORIZONTAL COMPONENT**WHAT ABOUT DIRECTION?**In the example, DISPLACEMENT asked for and since it is a VECTOR quantity, we need to report its direction. N W of N E of N N of E N of W N of E E W S of W S of E NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. W of S E of S S**NEED A VALUE – ANGLE!**Just putting N of E is not good enough (how far north of east ?). We need to find a numeric value for the direction. To find the value of the angle we use a Trig function called TANGENT. 200 km 160 km, N q N of E 120 km, E So the COMPLETE final answer is : 200 km, 53.1 degrees North of East**What are your missing components?**Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components? The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions since and cosine. H.C. = ? V.C = ? 25 65 m**For tonight, find the following vertical and horizontal**components**Example**A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement. 23 m, E - = 12 m, W - = 14 m, N 6 m, S 20 m, N 14 m, N 35 m, E R q 23 m, E The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST**Example**A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. 8.0 m/s, W 15 m/s, N Rv q The Final Answer : 17 m/s, @ 28.1 degrees West of North**Example**A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components. H.C. =? 32 V.C. = ? 63.5 m/s**Example**A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement. 1500 km V.C. 40 5000 km, E H.C. 5000 km + 1149.1 km = 6149.1 km R 964.2 km q The Final Answer: 6224.2 km @ 8.92 degrees, North of East 6149.1 km**Homework #1**• Remember YOU CANNOT SOLVE A VECTOR PROBLEM WITHOUT DRAWING A PICTURE!!!!! Use lots of space. • Honors Physics: P23 Q21 – 26 • Regular Physics: P91 Q1,2,3 and P94 Q1,2,3,4,5,6,7**Adding vectors algebraically that are not at right angles:**• So far the vectors have been “tidy” and at right angles to each other. In real life that is rarely the case. • Consider this problem: • A 200N force is placed on a box 15° W of N. Another 150N force is placed on the box 25° E of N. Find the net horizontal force on the box.**Strategy for dealing with adding vectors:**• Draw an accurate picture of all vectors. • Add and /or subtract collinear vectors to reduce the problem. • Find x and y components of each remaining vector. Be consistent with + and – signs. • Add all x components for all vectors. • Add all y components for all vectors • Reconstruct the resultant vector from x and y component sums, STATING MAGNITUDE AND ANGLE IN ANSWER. .**A 200N force is placed on a box 15° W of N. Another 150N**force is placed on the box 25° E of N. Find the net horizontal force on the box.**Projectiles – anything shot or thrown into the air that**have only gravity as a net external force on them**Projectiles**• Any object that is shot or thrown into the air. • Examples: Baseball, bullet, basketball, tennis ball, etc. can you think of others. • The following are not projectiles: Rockets, aircraft, birds, helicopters etc. can you think why? • The path of a projectile throught the air is called the TRAJECTORY. • Projectiles may be horizontal or launched at an angle.**Horizontal Projectiles**• Examples: a ball rolled of the edge of a table, a car driving off a cliff etc. Can you think of any more? • The time it takes a horizontal projectile to reach the ground is the same time taken for an object to fall straight down from the same height. • The range (how far it lands from the base of cliff etc) depends on its speed at time of launch and the height of the drop. • Let’s do the following problem...**Projectiles launched at an angle**• Examples would be throwing a ball far, throwing a javelin. Can you think of more? • These projectiles have a horizontal and vertical speed at the launch. • The range depends on speed of launch and angle alone. • The angle for maximum range is 45 degrees • Angles > 45 make it stay in the air longer. • Angles < 45 make it land sooner.