Lesson 3 Kinematics.pptx

KINEMATICS OF TRANSLATIONAL MOTION PHYSICS LESSON 3  KINEMATICS OF TRANSLATIONAL MOTION Kinematics Quantities Instantaneous and Position Average Acceleration https://upload.wik...

KINEMATICS OF TRANSLATIONAL MOTION PHYSICS LESSON 3  KINEMATICS OF TRANSLATIONAL MOTION Kinematics Quantities Instantaneous and Position Average Acceleration https://upload.wikimedia.org/wikipedia/commons/thumb/4/40/ https://science4fun.info/wp-content/uploads/2017/02/ Distancedisplacement.svg/300px-Distancedisplacement.svg.png accelerating-van.jpg 3.1 KINEMATIC QUANTITIES POSITION  (INSERT PICTURE SPECIFYING A POSITION RELATIVE TO A CONVENIENT REFERENCE FRAME) https://noschese180.files.wordpress.com/2014/09/positionnumberline 3.1 KINEMATIC QUANTITIES POSITION  Time  Distance vs. Displacement  Displacement  Speed vs. Velocity  Velocity  Average Velocity TIME  Time is the change or the interval over which change occurs.  Second (s) – SI unit for time  Elapsed time is the difference between ending time and beginning time: t = tf – to Where: t = change in time / elapsed time tf = ending time t = beginning time TIME = 6:30 – 3:00 = 3:30 or 3.5 hours Onlinemathlearning.com  John is preparing for a hiking trip to a nearby mountain. He plans to start hiking at 9:30 AM and expects to reach the mountain's summit at 12:45 PM. During his hike, he also wants to take a 15-minute break at a scenic viewpoint.  How long will it take John to reach the mountain's summit without considering the break?  What time does John expect to finish his hike and return to his starting point if he spends 15-minute break and additional 30 more minutes exploring the summit?   How long will it take John to reach the mountain's summit without considering the break?  Given:  =12:45PM  =9:30Am = 12:45 – 9:30 = 3hrs and 15 mins or 3.15 hours  What time does John expect to finish his hike and return to his starting point if he spends 15-minute break and additional 30 more minutes exploring the summit?  Let's break it down:  John plans to start hiking at 9:30 AM.  He expects to reach the mountain's summit at 12:45 PM.  He plans to take a 15-minute break at a scenic viewpoint.  He also wants to spend an additional 30 minutes exploring the summit.  After exploring the summit, he will need to hike back down to his starting point.  What time does John expect to finish his hike and return to his starting point if he spends 15-minute break and additional 30 more minutes exploring the summit?  Time spent from starting point to the summit is equal to 3.15hrs.  Time spent from the summit going = 12:45 – 9:30  back to his starting point to the summit is equal to 3.15hrs. = 3hrs and 15 mins  Total duration back-and-forth is equal to 6hrs and 30 min.  Total duration back-and-forth + 15min + 30min =7hrs and 15min  = 7hrs and 15min  (9:30AM) + 7hrs and 15 minutes 4:45PM End Time DISTANCE VS. DISPLACEMENT  Distance- magnitude or size of displacement between two positions - distance traveled is the total length of the path traveled between two positions  Displacement- defined by both direction and magnitude DISTANCE VS. DISPLACEMENT DISPLACEMENT  It is the change in position of an object x = x f – x0  Where: x = displacement xf =final position x0 = initial position  DISPLACEMENT Given: xf = 3.5 m x0= 1.5 m Required: x ) Solution: x = xf – x0 = 3.5 m – 1.5 m = 2.0 m khanacademy.org  Maria is taking a walk in her neighborhood. She starts her walk from her house, which she considers her initial position. She walks 200 meters east to visit a friend's house, stays for a while, and then walks 150 meters west to reach the neighborhood park. She notes her final position at the park. Maria's initial position at her house is considered (0 meters) since it's her starting point.  What is her final position at the park?  Calculate Maria's displacement, which represents the change in her position from her house to the park.  What is her final position at the park?  Given:  x1=200m to East  x2=150m to West x = x f – x0 xf= x0 + (Distance east) - (Distance west) xf= x0 + 200m - 150m xf= 50m E  Calculate Maria's displacement, which represents the change in her position from her house to the park.  Given:  xf= 50m E  x0=0m x = x f – x0 x = x f – x0 x = 50m – 0 m x = 50m E SPEED VS. VELOCITY  Speed has no direction (scalar quantity)  Velocity has units of distance divided by time, such as mph or km/h AVERAGE VELOCITY  It is the displacement (change in position) divided by the time of travel. ( or vave) =  Where: = average velocity = change in position or displacement ( are the final and beginning positions) = change in time ( are the final and beginning time) AVERAGE VELOCITY =  An airplane passenger took 5 seconds to move -4m horizontally. His average velocity would be: = = W AVERAGE VELOCITY  Given: You travel 370 km west to a friend’s house. You left at 10 x = 370 km am and arrived at 3pm. What was your average velocity? to = 10am tf = 3pm  Required: or vave  Solution: = = = -74 km/h, west 3.2 AVERAGE AND INSTANTANEOUS ACCELERATION In everyday conversation, to accelerate means to speed up. The greater the acceleration, the greater the change in velocity over a given time. clipartkey.com 3.2 AVERAGE AND INSTANTANEOUS ACCELERATION Average Acceleration  Acceleration as a Vector  Deceleration vs. Negative Acceleration  Instantaneous Acceleration dreamstime.com 3.2 AVERAGE AND INSTANTANEOUS ACCELERATION Acceleration- a vector in the same direction as the change in velocity. pinterest.com AVERAGE ACCELERATION  Average acceleration is the rate at which velocity changes =  Where: = average acceleration v = velocity t = time AVERAGE ACCELERATION  The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s going to north. What is the car’s average acceleration? = = = = = 4 m/s2 N  Tom is riding his bicycle. He starts from rest and accelerates uniformly to a speed of 12 meters per second (m/s) in 4 seconds going to south. Calculate Tom's average acceleration during this time.  Sarah is taking a road trip from City A to City B, which are 200 miles apart. She starts her journey at 8:00 AM and arrives in City B at 10:00 AM. During this time, she makes a rest stop for 30 minutes at a diner along the way. Calculate Sarah's average velocity for the entire trip, considering the time she spent driving and resting. ACCELERATION AS A VECTOR ACCELERATION VELOCITY Change in speed Change in magnitude Change in direction  DECELERATION VS. NEGATIVE ACCELERATION DECELERATION NEGATIVE ACCELERATION  It refers to the acceleration  Acceleration in the negative in the direction opposite to direction in the chosen the direction of the velocity. coordinate system.  Always reduces in speed gfycat.com nroer.gov.in DECELERATION VS. NEGATIVE ACCELERATION QUESTION:  Can deceleration be negative acceleration?  Can negative acceleration be deceleration? INSTANTANEOUS ACCELERATION  The acceleration (a) at a specific instant in time, is obtained by the same process as discussed for instantaneous velocity.