Vector and Scalar Quantities PDF

Vector and Scalar Quantities LESSON 2 –GENERAL PHYSICS 1 JOVIT A. HAL AS AN, LPT Objectives:  Differentiate vector and scalar quantities.  Perform addition of vectors.  Rewrite a vector in component form.  Calculate directions and magnitudes of vectors. On the map, we can see diff...

Vector and Scalar Quantities LESSON 2 –GENERAL PHYSICS 1 JOVIT A. HAL AS AN, LPT Objectives:  Differentiate vector and scalar quantities.  Perform addition of vectors.  Rewrite a vector in component form.  Calculate directions and magnitudes of vectors. On the map, we can see different physical quantities such as distance in meters or in kilometers together with the specific directions. Vector and Scalar Quantities Vector Quantity Is a physical quantity that has both magnitude and direction. Examples: Displacement of 20m (North), Velocity of 60 km/h (East) Force of 40 Newton (West) Scalar Quantity Is a physical quantity that has only magnitude. Example: Speed of 60km/h Time of 3hours Temperature of 28˚C SCALAR VECTOR SCALAR VECTOR SCALAR VECTOR PRACTIC E PRACTICE#1 Determine whether the following each quantity is Scalar or Vector. 50 minutes Scalar PRACTICE#2 75 minutes due East Vector PRACTICE#3 82 Newtons to the left Vector PRACTICE#4 14 m/s, west Vector PRACTICE#4 36 Scalar SCALAR ADDITION and SUBTRACTION SCALAR ADDITON & SUBTRACTION Adding scalar quantities is similar to ordinary ad dition. We add together the quantities express in the same units. Example: mass (m1)=25g and another mass (m2)=50g their sum is msum = m1 + m2 Vector Addition & Subtraction Vectors We always draw a vector as a line with an arrowhead at its tip. The length of the line shows the vector’s magnitude, and the direction of the line shows the vector’s direction. Displacement Displacement represents a vector quantity such as displacement by a single letter, This symbol is in boldface italic type with an arrow above them, to represent that it have different properties from scalar quantities; the arrow is a reminder that vectors have direction. Displacement Is simply a change in position of an object. State not only how far the object moves but also in what direction. Walking 3 km north from your front door doesn’t get you to the same place as walking 3 km southeast; these two displacements have the same magnitude but different direction. Displacement Displacement is always a straight- We represent a displacement by an arrow line segment directed from the pointing in the direction of displacement. starting point to the ending point, even though the object’s actual Ending position: path may be curved. If two vectors have the same direction, they are parallel. Displacement If they have the same magnitude and the same direction, they are Starting position: equal, no matter where they are located in space. Displacement The vector ’ from point to point has the same length and direction as the vector 𝑃2 𝑃4 from point to. These two displacements are equal, even though they start at different points. ⃗ 𝑨 ’ We write this as ’ = ; the boldface equal sign emphasizes that equality of two ’ = vector quantities is not the same relationship as equality of two scalar 𝑃1 𝑃3 quantities. Two vector quantities are equal only when they have the same magnitude and same direction. Displacement The vector , however, is not equal to because its direction is opposite 𝑃2 𝑃5 to that of. We define the negative of a vector as a vector having the same magnitude as the original vector but ⃗ 𝑨 the opposite direction. - The negative of vector quantity is denoted as and we use a boldface 𝑃1 𝑃6 minus sign to emphasize the vector nature of the quantities. Displacement If is 87 m north, then is 87 m south. Thus, we can write the relationship 𝑃2 𝑃5 between and as or = When two vectors and have opposite ⃗ 𝑨 directions, whether their magnitudes are the same or not, we say that they are antiparallel. 𝑃1 𝑃6 Vector addition Suppose a particle undergoes a displacement followed by a second ⃗ 𝑩 displacement. The final result is the same as if the ⃗ 𝑨 particle had started at the same initial ⃗ 𝑪 point and undergone a single displacement. We call displacement the vector sum, or resultant, of displacements and. We express this relationship symbolically as: Vector addition If we make the displacement and in reverse order, with first and second, the results is the same ⃗ and + = + 𝑪 ⃗ 𝑨 This shows that the order of terms in a vector sum doesn’t matter. In other words, vector addition ⃗ obeys the commutative law. 𝑩 Example#1: = 2 units, East = 4 units, East Calculate the resultant Example#2: = 2 units, East = 4 units, West Calculate the resultant. Example#3: = 80 units, North = 120 units, South Calculate the resultant. Example#4: = 30 units, East = 40 units, North Calculate the magnitude of the resultant resultant vector and find the direction. Example#5: = 50 units, west = 120 units, south Calculate the magnitude of the resultant resultant vector and find the direction. Example#6: = 45 units, East = 60 units, south Calculate the magnitude of the resultant resultant vector and find the direction. Note: = Can be calculated through inverse tangent = = = Vector Addition: Head-to-Tail- Method A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1˚ north of east. Vector Addition: Head-to-Tail- Method Draw an arrow to represent the y total displacement vector D. Using protractor, draw a line at an angle relative to the east-west axis. The length D of the arrow is proportional to the vector’s i ts magnitude and is measured along u n 0.3 the line with a ruler. 1 In this example, the magnitude 𝜃=29.1 ˚ D of the vector is 10.3 units, x and the direction is 29.1˚ north of east. Vector Addition: Head-to-Tail- Method Step 1. Draw an arrow to represent the first vector (9 y blocks to the east) using a ruler or protractor. a 𝜃=0 ˚ x 9 units Vector Addition: Head-to-Tail- Method Step 2. Now draw an arrow to represent the second vector (5 y blocks to the north). b Place the tail of the second vector at the head of the first vector. x Vector Addition: Head-to-Tail- Method Step 3. If there are more than two vectors, continue this process y for each vector to be added. c Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail. Step 4. Draw an arrow from the.3 tail of the vector to the head of 10 ts i un the last vector. This is the 𝜃=29.1 ˚ resultant, or the sum, of the other vectors. x Vector Addition: Head-to-Tail- Method Step 5. To get the magnitude of the resultant, measure its length y with a ruler. c (Note that in most calculations, we will use the Pythagorean theorem to determine this length)..3 Step 6. To get the direction of the 10 ts i un resultant, measure the angle it 𝜃=29.1 ˚ makes with the reference frame using a protractor. x PRACTIC E PRACTICE#1 = 5 km, East = 7 km, North Calculate the magnitude of the resultant resultant vector and find the direction. PRACTICE#2 A car covered 25 km, 60 north of east on its initial route. Afterwards, it covered 50 km in the direction 30 north of west. Using the analytical method, what is its resultant displacement and direction? PRACTICE#3 Matthew leaves the base camp and hikes 11km, north and then hikes11km east. Determine Matthew's resulting displacement and direction. PRACTICE#4 Matthew leaves the base camp and hikes 11km, north and then hikes11km east. Determine Matthew's resulting displacement. PRACTICE#5 Jomar rode his bicycle in the eastern direction for 5 meters. After that, he went 3 meters in a direction of 30° North of East. Determine Jomar’s displacement.  PT#1 MATERIALS NEEDED: PROTRACTOR, METERSTICK/TAPE MEASURE, NILON OR SEWING THREAD, FOLDER SLIDER, AND METAL RING. Measure the height of the flag pole  Resultant of Two Vectors Using Cosine Law and Sine Law 2 2 2 𝑐 =𝑎 + 𝑏 − 2 𝑎𝑏𝑐𝑜𝑠𝐶 B c A C 𝒂 = 𝒃 Example#1 = 50 m, west = 30m, 60 degrees to the North East Calculate the magnitude of the resultant vector and find the direction. Example#2 = 80 m, east = 120m , 135 degrees to the Northwest Calculate the magnitude of the resultant vector and find the direction. Example#3 ni t s rees eg 40 u d 50 n its u 30 Calculate the magnitude of the resultant vector and find the direction. Component of Vectors Representing a vector in terms of (a) component vectors and and (b) components and (which in this case are both positive). To define what we mean by the y components of vector we begin with a rectangular (Cartesian) coordinate system of axes. The component vector of We then draw the vector with its tail at 0, the origin of the coordinate system. We can represent any vector lying in the xy- ⃗ 𝐴𝑦 plane as the sum of a vector parallel to the ⃗ 𝑨 x-axis and a vector parallel to the y-axis. These two vectors are labeled and they are called the component vectors of 𝜃 vector , and their vector sum is equal to. x In symbols, 0 = +  Since each component vector lies along a coordinate-axis direction, we need only a single number to Component of Vectorsdescribe each one. When points in the positive x- y direction, we define the number to be equal to the magnitude of The component vector of When points in the negative x- direction, we define the number to ⃗ be equal to the negative of that 𝐴𝑦 ⃗ 𝑨 magnitude (the magnitude of a vector quantity is itself never negative). 𝜃 We define the number in the same 0 x way. The two numbers and are called the components of.  We can calculate the components of the vector if we know the magnitude Component of Vectorsand its direction. We’ll describe the direction of a vector y by its angle relative to some reference direction. In the figure, this reference direction is the positive x-axis, and the The component vector of angle between vector and the positive x-axis is. ⃗ 𝑨 Imagine that the vector originally lies 𝐴 𝑦 = 𝐴𝑠𝑖𝑛 𝜃 along the +x-axis and that you then rotate it to its correct direction, as 𝜃 indicated by the arrow on the angle. 0 𝐴𝑥 = 𝐴𝑐𝑜𝑠 𝜃 x  If this rotation is from the +x-axis toward the +y-axis, then is positive; if the rotation is from the +x-axis Components of Vectors toward the –y-axis, is negative. Thus, the +y-axis is at an angle of y 90˚, the –x-axis at 180˚, and the –y- axis at 270˚ (-90˚). The component vector of If is measured in this way, then from the definition of the trigonometric function, ⃗ 𝑨 and 𝐴 𝑦 = 𝐴𝑠𝑖𝑛 𝜃 Eq. 6 𝜃 and 0 𝐴𝑥 = 𝐴𝑐𝑜𝑠 𝜃 x ( measured from the +x-axis, rotating toward the +y-axis)  Components of Vectors In the figure and are positive. This y is consistent with the equations; is in the first quadrant (between 0˚ The component vector of and 90˚), and both the cosine and sine of an angle in this quadrant are positive. ⃗ 𝑨 𝐴 𝑦 = 𝐴𝑠𝑖𝑛 𝜃 𝜃 0 𝐴𝑥 = 𝐴𝑐𝑜𝑠 𝜃 x  The components of a vector may be positive or negative numbers y is positive: Its component vector points in the +y- direction ⃗ 𝑩 𝐵 𝑦 ¿ In this figure, the component is negative. 𝜃x Again, this agrees with the equations; the 𝐵 𝑥 ( −) cosine of an angle in the second quadrant is is negative: Its component negative. The component is positive. vector points in the –x- direction 𝐶 𝑥 ( −) x ⃗ 𝐶 𝑦 (−) 𝑪 𝜃 Both components of are negative y EXAMPLE#1 (a) What are the x- and y- y components of vector in the figure? The magnitude of the vector is D=3.00 m, and the angle. 𝐷𝑥 ¿ x 𝑎 𝐷 𝑦 (−) ⃗ 𝑫 EXAMPLE#2 (b) What are the x- and y- components of vector in x y the figure? The magnitude of the vector is E=4.50 m, 𝐸𝑥¿ and the angle β. 𝐸 𝑦 (−) β ⃗ 𝑬 Example#3 = 100 units, East = 150 units, 30 Northeast Calculate the magnitude of the resultant resultant vector and find the direction. CALCULATING THE X AND Y COMPONENTS OF VECTORS Given: R = 50; 𝜽 = 23 y ⃗ 𝑹=𝟓𝟎 𝒎 @ 𝜽=𝟐𝟑 ⃗𝑹 𝐴𝑦 ¿ Solution: 𝜃x 𝐴 𝑥 (−) = Rcos𝜽 = Rsin𝜽 = 50cos(23 = 50sin(23) = 46.03m = 19.6m  FINDING COMPONENTS OF VECTORS \ y y 𝐴 𝒊 𝒕𝒔 𝐴 𝑦 𝑦 𝒖𝒏 𝟐𝟓 𝟏𝟐 𝒎 60° 30 ° 0 𝐴𝑥 0 𝐴𝑥 x x  PRACTICE#1 y 𝐴 𝟔 𝒎 𝟑 𝑦 50.4° 0 x 𝐴𝑥 Example#4 Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacement) on a flat field. First she walks 25.0 m in a direction 49.0˚ north of east. Then, she walks 23.0 m heading 15.0˚ north of east. Finally, she turns and walks 32.0 m in a direction 68.0˚ south of east. Calculate the magnitude of the resultant resultant vector and find the direction. PRACTIC E PRACTICE#1 = 85 meters, west = 130 units, 40 Northeast Calculate the magnitude of the resultant resultant vector and find the direction. PRACTICE#2 During his early morning training, Louie jogged 10 km, 20° south of west. He then covered another 15 km in the direction of 60° south of east before resting. What is his resultant displacement? PRACTICE#3 Denise walks every day from her house to the school. First, she covers 10 m, 20° north of east. Then, she walked 15 m in a direction 50° north of east. What is her resultant displacement? PRACTICE#4 Find the resultant vector R if A is 95 N, 30° north of east, B is 50 N, south while C has a magnitude of 75 N and a direction of 45° south of west? KINEMATICS: MOTION ALONG A STRAIGHT LINE LESSON 3 –GENERAL PHYSICS 1 JOVIT A. HAL AS AN, LPT  Acceleratio Displacement VS. Distance n Vs. Deceleratio n Speed vs. velocity Constant velocity Vs. Velocity vs. Constant Acceleration Acceleration  KINEMATICS  Describes motion in terms of displacement, velocity, and acceleration. Motion  Is the displacement of an object in relation to objects that are considered to be stationary.  It can also be defined as the continuous change of position with respect to a certain reference point.  DISTANCE AND DISPLACEMENT Distance Vs. Displacement Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position. △  THE DIRECTION OF THE DISPLACEMENT VECTOR Example#1 Calculate the displacement of a car as it moves from position 1 towards position 2. Answer: △x = xf - xi = -5 - 0 = -5  THE DIRECTION OF THE DISPLACEMENT VECTOR Example#2 Calculate the displacement of a car as it moves from position 2 towards position 3. Answer: △x = xf - xi = 5 – (-5) = 10  THE DIRECTION OF THE DISPLACEMENT VECTOR Example#3 Calculate the displacement of a car as it moves from position 3 towards position 2. Answer: △x = xf - xi = -5 – 5 = -10  EXAMPLES PROBLEM Example#4 A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car.  EXAMPLES PROBLEM Example#5 A runner travels around rectangle track with length = 50 meters and width = 20 meters. After travels around rectangle track two times, runner back to starting point. Determine distance and displacement. PRACTICE#1  SPEED AND VELOCITY Speed Vs. Velocity Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. Where d= distance t = time  SPEED AND VELOCITY Speed Vs. Velocity Velocity is a vector quantity that refers to "the rate at which an object changes its position.“  SPEED AND VELOCITY Example#1: Jefrey walks 200 m from his house to school. Upon his arrival to the gate of the school, he just realized that he needs to go to a store which is halfway from his house to buy a pen. What are his a) average speed and; b) average velocity when he arrived at the store if the time it takes for the entire trip INSTANTANEOUS SPEED AND VELOCITY INSTANTANEOUS SPEED AND VELOCITY  INSTANTANEOUS SPEED AND VELOCITY  Instantaneous speed is actually the magnitude of the instantaneous velocity. Given the value of the instantaneous speed, you just add its direction then it will now become the instantaneous velocity of an object.  AVERAGE ACCELERATION  Average acceleration is defined as the rate of change in velocity over time.  An object is accelerating if the change of its velocity is in an increasing manner while an object is decelerating if the change of its velocity is in a decreasing manner.  INSTANTANEOUS SPEED AND VELOCITY EXAMPLE#1 A small car was seen overtaking a truck that was travelling at a constant speed. This car started at 40 kph and was moving at 65 kph after 5 minutes. The car and the truck were then running side by side for 10 minutes. Then, the car left the truck behind as it sped up to 70 kph in 5 minutes. A. What was the average acceleration of the small car in the first 5 minutes?  INSTANTANEOUS SPEED AND VELOCITY EXAMPLE#1 A small car was seen overtaking a truck that was travelling at a constant speed. This car started at 40 kph and was moving at 65 kph after 5 minutes. The car and the truck were then running side by side for 10 minutes. Then, the car left the truck behind as it sped up to 70 kph in 5 minutes. B. What was the car’s average acceleration in the last 5 minutes?