Module 2- KINEMATICS Marlon Flores Sacedon Visayas State University Department of Physics Visca, Baybay City, Leyte Kinematics of Motion in One-Dimension QUESTIONS: o How long will it take to travel? o What is the distance traveled? o How fast does it move? o Etc. Kinematics of Motion in One-Dimension Topics o Introduction to kinematics o Uniformly Accelerated Bodies (UAB) o Freely-Falling Bodies (FFB) o Projectile Motion Kinematics of Motion in One-Dimension o Introduction to kinematics What is Kinematics? o A branch of classical mechanics that deals with the study of the motion of particle without considering the force that causes them to move o A purely descriptive study of motion o This will answers the question, how object moves? Kinematics of Motion in One-Dimension o Introduction to kinematics Why study Kinematics? Kinematics is a significant field of mechanics that helps us understand how objects move. By understanding the concepts involved in kinematics, we can predict an object's future motion and solve related problems. Kinematics of Motion in One-Dimension o Introduction to kinematics How to describe motion? β Time (π‘) β Position (π₯) β Velocity (π£) β Acceleration (π) β π₯π‘, π£π‘, and ππ‘ Graphs Kinematics of Motion in One-Dimension o Introduction to kinematics Definition of terms β Displacement is a change in position βπ₯ = π₯π − π₯π β Velocity is a change in position per time interval βπ₯ π£ππ£π = βπ‘ ππ₯ π£πππ = ππ‘ β Acceleration is a change in velocity per time interval βπ£ πππ£π = βπ‘ ππ£ ππππ = ππ‘ Kinematics of Motion in One-Dimension o Introduction to kinematics Examples of π₯π‘, & π£π‘- Graphs π₯π‘- Graphs vπ‘- Graphs Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies What is Uniformly Accelerated Bodies? Uniformly Accelerated Bodies (UAB) is a study of motion of particle at constant acceleration. ππ£ ππππ = ππ‘ ππ£ π= ππ‘ Acceleration is constant Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies Six Physical Quantities of UAB 1. Time interval, π‘ ------------ (π ) 2. Displacements, s ---------- (π) 3. Average velocity, π£ππ£π ---- (π/π ) 4. Initial velocity, π£π ---------- (π/π ) 5. Final velocity, π£ ------------ (π/π ) 6. Acceleration π ------------- (π/π 2 ) Six Formulas of UAB π 1) π£ππ£π = π‘ π£+π£0 2) π£ππ£π = 2 π£−π£0 3) π = π‘ 5) π = π£0 π‘ + 1Τ2 ππ‘ 2 4) π£ = π£0 + ππ‘ 6) π£ 2 = π£0 2 + 2ππ Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies PROBLEMS in UAB 1) Given: π = 10 m ; π‘ = 3 π ; π = 2 m/s2 Required: π£π , π£, π£ππ£π Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies PROBLEMS in UAB 2) Given: π = 150 cm ; π‘ = 2 π ; π£π = 14 cm/s Required: π, π£ππ£π , π£ Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies Quiz #2 (5 pts) Given: π = 50 m ; π = 2m/s2; π£π = 9 m/s Required: t, π£ππ£π , π£ Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies Problem: An antelope moving with constant acceleration covers the distance between two points 60.0 π apart in 5.00 π . Its speed as it passes the second point is 14.0 π/π . What are (a) its speed at the first point and (b) its acceleration? Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies Applications PROBLEM: A bird is flying due east. Its distance from a tall building is given by π₯ π‘ = 28.0π + 12.4π/π π‘ − (0.045 π/π 3 )π‘ 3 . What is the instantaneous velocity of the bird when π‘ = 8.00 π ? Kinematics of Motion in One-Dimension o Uniformly Accelerated Bodies Try this: The graph in Figure shows the velocity of a motorcycle police officer plotted as a function of time. (a) Find the instantaneous acceleration at π‘ = 3 π , π‘ = 7 π , and π‘ = 11 π . (b) How far does the officer go in the first 5 π ? The first 9 π ? The first 13 π ? Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) βͺ Freely falling bodies is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. βͺ Air resistance and other external forces are neglected. βͺ In the absence of air resistance, regardless its size and shape, the body will fall at the same rate. βͺ Free-fall is a good example of Uniformly Accelerated Bodies (UAB). Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) Gravitation acceleration 22 +π¦ π = −9.81 9.81 π/π π/π +π¦ 0 0 −π¦ −π¦ Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) Six Physical Quantities of FFB 1. Time interval, π‘ ------------ (π ) 2. Displacements, π¦ ---------- (π) 3. Average velocity, π£ππ£π ---- (π/π ) 4. Initial velocity, π£ππ¦ ---------- (π/π ) 5. Final velocity, π£π¦ ------------ (π/π ) 6. Gravitational Acceleration π = −9.81 π/π 2 Six Formulas of UAB π 1) π£ππ£π = π‘ 2) π£ππ£π = 3) π = π£π¦ +π£0π¦ 2 π£π¦ −π£0π¦ π‘ 4) π£π¦ = π£0π¦ + ππ‘ 5) π¦ = π£0π¦ π‘ + 1Τ2 ππ‘ 2 6) π£π¦ 2 = π£0π¦ 2 + 2ππ Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) PROBLEM: A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands on the street below. Ignore air resistance. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street? Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) PROBLEM: You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance. Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) PROBLEM: You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling? Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) 1) You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance. 2) A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell? 3) A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands on the street below. Ignore air resistance. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street? Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) Problems 1) (a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air? 2) A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand? Kinematics of Motion in One-Dimension o Freely-falling Bodies (FFB) Freely-falling Bodies (FFB) ASSIGNMENT 1) (a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air? ANSWER: 2) A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn’s satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus? ANSWER: 3) A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand? ANSWER: Projectile Motion The Idealized Model Kinematics on 2D or 3D Motion of Point Particle QUESTION # 1: How many possible angle/s of projection in order to shot the ball into the cylinder’s opening? QUESTION #2: What are the magnitudes of these angles? 12 π/π π2 = ? 2π π1 = ? 7π CANNON TARGET OPEN CYLINDER To answer… Study Projectile Motion Learning Outcomes Derive the formulas for Idealized model of Projectile Motion and apply to real life problems. 6. π₯ = π£ππ₯ β π‘ 1. π£ππ₯ = π£π cos π 1 2 7. π¦ = π£ππ¦ β π‘ − 2ππ‘ 2. π£ = π£ sin π ππ¦ π 3. π£π₯ = π£ππ₯ 4. π£π¦ = π£ππ¦ − ππ‘ 5. π£ = π£π₯ 2 + π£π¦ 2 8. π₯ π‘= π£π cos π ππ₯ 2 π¦ = π₯ tan π − 2π£ 2cos2π π Learning Outcomes Derive the formulas for Idealized model of Projectile Motion and apply to real life problems. Tricky problem Recall: β Kinematics of Motion β Describing motion of particle/s β Motion β Change in positions β Position, time, velocity, & acceleration β Kinematics in One-Dimensional Motion o Uniformly accelerated bodies o Freely Falling Bodies Recall: β Some Trigonometric Identities o 1 cos π = 2 sec π 2 2 2 o sec π = 1 + tan π 1 cos2 π o sec 2 π = o sinπ tanπ = cosπ What is Projectile Motion? Trajectory +π¦ π₯, π¦, π‘ π£π π π¦ +π₯ π₯ Projectile Motion Time of Flight Initial Velocity of projected body +π¦ Angle of projection π‘ Position y π₯, π¦, π‘ π£π Position π₯ π +π₯ π₯ π¦ Projectile Motion: The Idealized Model ASSUMPTIONS: β Air resistance & other effects are negligible +π¦ π‘ π₯, π¦, π‘ π£π π¦ π +π₯ π₯ βͺ Along π₯-axis: Uniform motion ππ = π βͺ Along y-axis: Free-fall ππ = π = −π. ππ m/s2 Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π£π¦ = 0 π£π¦ +π¦ π£ππ¦ π£ π£π₯ π£π π π£ππ₯ π£ π£π₯ π£π₯ π₯, π¦, π‘ π = −9.81 π£π¦ π£ π¦ +π₯ π₯ βͺ Along π₯-axis: Uniform motion ππ = π βͺ Along y-axis: Free-fall ππ = π = −π. ππ m/s2 Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible +π¦ Let’s derive the formulas π‘ π₯, π¦, π‘ π£ππ¦ π£π π£π π£ππ¦ π£ππ₯ π = −9.81 π¦ π +π₯ π₯ π Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible +π¦ π‘ π₯, π¦, π‘ π£ππ¦ π£π π = −9.81 π¦ π π£ππ₯ π£ππ₯ +π₯ cos π = π£π π₯ 1. π£ππ₯ = π£π cos π π£ππ₯ = π£π cos π π£π π π£ππ₯ π£ππ¦ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible +π¦ π‘ π₯, π¦, π‘ π£ππ¦ π£π π = −9.81 π¦ π π£ππ₯ 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π π£ππ¦ +π₯ sin π = π£π π₯ π£ππ¦ = π£π sin π π£π π π£ππ₯ π£ππ¦ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π£π¦ +π¦ π£ π£π₯ π£ππ¦ π£π¦ π£π π‘ π₯, π¦, π‘ π = −9.81 π¦ π π£π£π₯ππ₯ 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π +π₯ π₯ ππ£ π= ππ‘ π‘π£ ΰΆ±: ππ£ = ΰΆ± = π ππ‘ 0 π£π π£ − π£π = ππ‘ π£ = π£π + ππ‘ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π£π¦ +π¦ π£ π£π₯ π£ππ¦ π£π π‘ π₯, π¦, π‘ π = −9.81 π¦ π π£ππ₯ 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π +π₯ π₯ 3. π£π₯ = π£ππ₯ π£π₯ = π£ππ₯ + 0 π‘ π£π₯ = π£ππ₯ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π£π¦ +π¦ π£ π£π₯ π£ππ¦ π£π π‘ π₯, π¦, π‘ π = −9.81 π¦ π π£ππ₯ 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π +π₯ π₯ 3. π£π₯ = π£ππ₯ 4. π£π¦ = π£ππ¦ − ππ‘ π£π¦ = π£ππ¦ + −π π‘ π£π¦ = π£ππ¦ − ππ‘ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π£π¦ +π¦ π£ π£π₯ π£ππ¦ π£π π‘ π₯, π¦, π‘ π = −9.81 π¦ π π£ππ₯ 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π +π₯ π₯ 5. π£ = 3. π£π₯ = π£ππ₯ 4. π£π¦ = π£ππ¦ − ππ‘ π£π₯ 2 + π£π¦ 2 ππ‘ π£ = π£π + ππ‘ ππ = π£ππ + ππ‘ ππ‘ ππ‘ ΰΆ±: ΰΆ± ππ = π£π + ππ‘ ππ‘ 0 1 2 π = π£π π‘ + 2ππ‘ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π = −9.81 Trajectory is a parabolic curve +π¦ π‘ π₯, π¦, π‘ π£ππ¦ π£π π¦ π π£ππ₯ 6. π₯ = π£ππ₯ β π‘ 1 2 7. π¦ = π£ππ¦ β π‘ − 2ππ‘ π₯ = π£ππ₯ β π‘ 1 2 +π₯ π¦ = π£ππ¦ β π‘ − 2ππ‘ π₯ π₯ 8. π‘ = π₯ = π£π cos π β π‘ π£π cos π π¦ = π£π sin π β π‘ − 12ππ‘ 2 Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π = −9.81 +π¦ π‘ π₯, π¦, π‘ π£ππ¦ π£π π¦ π π£ππ₯ +π₯ π₯ π₯ 8. π‘ = π₯ = π£π cos π β π‘ π£π cos π 1 2 π¦ = π£π sin π β π‘ − 2ππ‘ Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible π = −9.81 +π¦ π‘ π₯, π¦, π‘ π£ππ¦ π£π π¦ π π£ππ₯ +π₯ π₯ 8. π‘ = π₯ π£π cos π π₯ = π£π cos π β π‘ π₯ ππ₯12 π₯ π¦ = π₯ tan π − π¦ = π£π sin π β − π( )2 2 2 2 π£π cos π 2π£ cos π£π cos ππ π ππ₯ 2 π¦ = π₯ tan π − 2π£ 2cos2π π Time independent formula Projectile Motion: The Idealized Model ASSUMPTIONS: Air resistance & other effects are negligible FORMULAS 1. π£ππ₯ = π£π cos π 2. π£ππ¦ = π£π sin π 3. π£π₯ = π£ππ₯ 4. π£π¦ = π£ππ¦ − ππ‘ 5. π£ = π£π₯ 2 + π£π¦ 2 6. π₯ = π£ππ₯ β π‘ 7. π¦ = π£ππ¦ β π‘ − 1 2 ππ‘ 2 8. π₯ π‘= π£π cos π ππ₯ 2 π¦ = π₯ tan π − 2π£ 2cos2π π Time independent formula APPLICATION 1: An object will be projected at an angle of 65o above the horizontal surface at a speed of 13 m/s. Ignoring air resistance, (a) what will be the horizontal displacement of the particle 1.5 second after being released? (b) What will be velocity during 1.5 seconds of flight? (c) What will be the maximum height of the particle above the starting point? APPLICATION 2: o THE TRICKY PROBLEM o 1 cos π = 2 sec π 2 2 sec π = 1 + tan2 π 2 = 7 tan π − 1.669 sec 2 π 2 = 7 tan π − 1.669(1 + tan2 π) 3.669 = 7 tan π − 1.669tan2 π Let: π = πππ§ π½ 9.81(7)2 2 = 7 tan π − 2(12)2 cos2 π 1.669 2 = 7 tan π − cos2 π 3.669 = 7π§ − 1.669z 2 1.669z 2 − 7π§ + 3.669 = 0 z 2 − 4.19π§ + 2.2 = 0 TRICKY APPLICATION PROBLEM o o 1 cos π = 2 sec π 2 2 sec π = 1 + tan2 π z 2 − 4.19π§ + 2.2 = 0 4.19 ± (4.19)2 −4(1)(2.2) π§= 2(1) Let: π = πππ§ π½ π§1 = 0.615 π1 = 31.59π π§2 = 3.575 π2 = 74.37π ANSWERS Time of Flights 8. π₯ π‘= π£π cos π PRACTICE PROBLEM PROBLEM 1: A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball? PRACTICE PROBLEM PROBLEM 2: On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0° above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell’s initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity. PRACTICE PROBLEM PROBLEM 3: A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0o above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. PRACTICE PROBLEM PROBLEM 4: A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? PRACTICE PROBLEM PROBLEM 5: A rock is thrown with a velocity π£π , at an angle of ππ from the horizontal, from the roof of a building of height β. Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of ππ . Assignment PROBLEM 6: Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance. Answer: 2.57 m QUIZ APPLICATION 1: A tennis ball rolls off the edge of a table top 0.75 m above the floor and strikes the floor at a point 1.40 m horizontally from the edge of the table. Ignore air resistance. a). Find the magnitude of the initial velocity. b). Find the time of flight. c). Find the magnitude of the velocity of the ball just before it strikes the floor . Thank you… Next topic: Newton’s Laws of Motion
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