Kinematics Unit. Objectives for Kinematics Unit 4.1: The student will distinguish between the concepts of displacement and distance. 4.2: The student.

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Presentation transcript:

Kinematics Unit

Objectives for Kinematics Unit 4.1: The student will distinguish between the concepts of displacement and distance. 4.2: The student will explain the concept of speed mathematically and graphically. 4.3: The student will distinguish between the concepts of speed and velocity. 4.4: The student will analyze graphs depicting velocity versus time. 4.5: The student will mathematically and graphically evaluate the concept of acceleration. 4.6: The student will graphically evaluate the relationships among displacement, velocity, acceleration, and time. 4.7: The student will solve problems involving kinematics. 4.8: The student will solve problems using vectors. 4.9: The student will conceptually explain horizontal and vertical components of projectile motion. 4.10: The student will make calculations involving projectile motion.

Motion What is motion? How do we measure/know that something is moving? All motion must be measured relative to something else – We must choose a frame of reference – Sitting in the room, rotating about Earth’s Axis, Revolving around the sun, traveling through space… Usually we’ll use the ground. – THERE IS NO ABSOLUTE FRAME OF REFERENCE

Displacement Displacement is a change in position. We measure the displacement by comparing an objects starting location to its final location. Displacement = final position – initial position. ∆x = x f - x i Ex: A frog hops away from the river. When he starts his journey he is 2m from the river. After 3min he is 5m from the river. What is his displacement? – ∆x = x f – x i – ∆x = 5m – 2m = 3m Displacement is NOT the same as distance – Ex: track

Displacement Ex2: An apple falls from a tree 4m off the ground. It hits a man on the head 1m before it hits the ground. What is it’s displacement. (Assume up is positive and down is negative) ∆x = x f – x i  ∆x = 1m – 4m = -3m  Displacement can be positive or negative.

Problem: how to create the fastest car using the given materials Hypothesis: We believe that…. Design/Materials Test/Experiment…. Includes Data….. Calculate your V f Conclusion

Sin City Invasion Materials – Unlimited Paper Paper clips Tape Hot glue Rubber bands Popsicle sticks Poker chips ( wheels ) Wooden Skewers Straws Dimensions: Cannot exceed: 3.5 inches wide 12 inches long 250 grams

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Velocity The average velocity is displacement divided by time. – v avg =∆x/∆t = (x f -x i )/(t f -t i ) – Units for v are m/s – Avg. v can be + or – depending on the displacement This is an average velocity. It does not mean the object traveled at this speed constantly, only that this was the average.

Velocity Examples Jessica runs from the start line to the finish of the 100m dash in 12.9s. What is her v avg ? You walk with an average v of 1.2m/s to the north for 9.5min. How far do you go? Simpson drives with a v avg = 48 km/h. How long will it take him to go 144km? v avg =∆x/∆t  V avg = (100m-0m)/(12.9s-0s)  V avg = 7.75 m/s v avg =∆x/∆t  1.2m/s = ∆x/9.5min  1.2m/s = ∆x/570s  ∆x = 684m v avg =∆x/∆t  48km/h =144km/ ∆t  ∆t = 3h

Velocity vs. Speed Speed is distance traveled/time – Since distance & displacement are not the same, speed and velocity are not the same. On a graph of displacement vs time, the slope of the line is the same as the average velocity ( if it was distance v time the slope would be the speed ) Instantaneous velocity is an object velocity at a single point in time. – The speedometer in your car show you instantaneous velocity.

Classwork – DO THESE! 1.Juan runs from the start line to the finish of the 50m dash in 4.9s. What is his v avg ? 2.You walk with an average v of 3.9m/s to the north for 15.3min. How far do you go? 3.Jack drives with a v avg = 55 km/h. How long will it take him to go m? – Make sure you check your units

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Acceleration Acceleration is the rate at which velocity changes. – a avg = ∆v/∆t – Units = m/s/s or m/s 2 Velocity and acceleration can both be positive or negative – pg 51 - chart

Constant Acceleration Means that the velocity is changing at the same rate in each time segment With a constant acceleration, we can get some equations for velocity and displacement. Displacement – ∆x = ½(v i + v f )∆t – ∆x = ½a(∆t) 2 + v i ∆t Velocity – v f = a∆t + v i Final v after any displacement – v f 2 = v i 2 + 2a∆x

Ex: Jane pushes a stroller from rest with a constant accel. of.50 m/s 2. What its velocity after it has gone 4.75m? v f 2 = v i 2 + 2a∆x – v f 2 = (0m/s) 2 + 2(.50 m/s 2 )(4.75m) – v f 2 = 4.75 m 2 /s 2 – √(v f 2) = √(4.75 m 2 /s 2) – v f = 2.18 m/s

Ex An airplane starts from rest and undergoes a constant acceleration of 4.8 m/s 2 for 15s before takeoff. A) what is it’s speed at take off? B) How long must the runway be? A) v f = a∆t + v i – v f = (4.8 m/s 2 )(15s) + 0m/s – v f = 72 m/s B) ∆x = ½a(∆t) 2 + v i ∆t – ∆x = ½(4.8m/s 2 )(15s) 2 + (0m/s)(15s) – ∆x = 540m

CLASSWORK: Do These! 1.An airplane starts from rest and undergoes a constant acceleration of 9.0 m/s 2 for 21s before takeoff. A) What is it’s speed at take off? B) How long must the runway be? 2. Ust pushes a stroller from rest with a constant acceleration of.40 m/s 2. What its velocity after it has gone 650 cm?