VECTORS PRE-AP PHYSICS SCALAR A SCALAR quantity is any quantity in physics that has MAGNITUDE ONLY Number value with units Scalar Example Magnitude Speed 35 m/s Distance 25 meters Age 16 years VECTOR A VECTOR quantity is any quantity in physics that has BOTH MAGNITUDE and DIRECTION Vector Example Magnitude and Direction Velocity 35 m/s, North Acceleration 10 m/s2, South Displacement 20 m, East Vector quantities can be identified by bold type with an arrow above the symbol. V = 23 m/s NE Vectors are represented by drawing arrows The length and direction of a vector should be drawn to a reasonable scale size and show its magnitude 20 km 10 km VECTOR APPLICATION •ADDITION: When two (2) vectors point in the SAME direction, simply add them together. •When vectors are added together they should be drawn head to tail to determine the resultant or sum vector. •The resultant goes from tail of A to head of B. LET’S PRACTICE A man walks 46.5 m east, then another 20 m east. Calculate his displacement relative to where he started. + 46.5 m, E 66.5 m, E 20 m, E VECTOR APPLICATION SUBTRACTION: When two (2) vectors point in the OPPOSITE direction, simply subtract them. LET’S PRACTICE SOME MORE…. 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 GRAPHICAL METHOD Aligning vectors head to tail and then drawing the resultant from the tail of the first to the head of the last. GRAPHICAL VECTOR ADDITION A + B Step 1 – Draw a start point Step 2 – Decide on a scale Step 3 – Draw Vector A to scale Step 4 – Vector B’s tail begin at Vector A’s head. Draw Vector B to scale. Step 5 – Draw a line connecting the initial start point to the head of B. This is the resultant. NON CO-LINEAR VECTORS When two (2) vectors are PERPENDICULAR to each other, you must use the PYTHAGOREAN THEOREM LET’S PRACTICE A man travels 120 km east then 160 km north. Calculate his resultant displacement. FINISH the hypotenuse is called the RESULTANT 160 km, N VERTICAL COMPONENT c 2 a2 b2 c a2 b2 c resul tan t c 200km 120 160 2 2 S R T T A 120 km, E HORIZONTAL COMPONENT WHAT ABOUT DIRECTION? In the example, DISPLACEMENT is asked for and since it is a VECTOR quantity, we need to report its direction. N W of N N of E NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. E of N N of E N of W W E S of W S of E W of S E of S S DIRECTIONS • There is a difference between Northwest and West of North 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 N of E Tan opposite side 160 1.333 adjacent side 120 Tan1 (1.333) 53.1o 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? H.C. = ? V.C = ? 25˚ 65 m The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions sine and cosine. adjacent side hypotenuse adj hyp cos cosine opposite side hypotenuse opp hyp sin sine adj V .C. 65 cos 25 58.91m, N opp H .C. 65 sin 25 27.47m, E 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. - 12 m, W - = 6 m, S = 14 m, N R 14 2 232 26.93m 20 m, N 35 m, E 23 m, E 14 .6087 23 Tan 1 (0.6087) 31.3 Tan 14 m, N R 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. Rv 82 152 17 m / s 8.0 m/s, W 15 m/s, N Rv 8 Tan 0.5333 15 Tan 1 (0.5333) 28.1 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˚ 63.5 m/s V.C. = ? adjacent side cosine hypotenuse adj hyp cos opposite side sine hypotenuse opp hyp sin adj H .C. 63.5 cos 32 53.85 m / s, E opp V .C. 63.5 sin 32 33.64 m / s, 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 40 5000 km, E H.C. adjacent side hypotenuse V.C. adj hyp cos cosine opposite side hypotenuse opp hyp sin sine adj H .C. 1500 cos 40 1149.1 km, E opp V .C. 1500 sin 40 964.2 km, N 5000 km + 1149.1 km = 6149.1 km R 6149.12 964.2 2 6224.2 km 964.2 0.157 6149.1 Tan1 (0.157) 8.92 o Tan R 964.2 km The Final Answer: 6224.2 km @ 8.92 degrees, North of East 6149.1 km