1 Speed, Velocity, Acceleration What is speed, velocity and acceleration?

2 Speed (s) = distance (d) time (t) Speed is the distance traveled per unit of time. Each variable measured by units: Distance : meters (m), miles (mi) Time : seconds (s), hours (hr), minutes (min) Speed: meters per second (m/s), miles per hour (mi/hr), kilometers per hour (km/hr)

3 Pretend you are looking at your car's speedometer while you are driving. The reading you get from your speedometer is This is the speed that you are traveling at that moment. A. instantaneous speed… 3 TYPES OF SPEED Instantaneous, Average, Constant

4 C. Average speed is the total distance traveled divided by the total time. It can be calculated using the following formula: speed = distance time s = d t...or shortened: Average Speed is: Total distance = all distance traveled over Total time = final time (end) minus initial time (beginning ) B. Constant Speed is when the object covers equal distances in equal amounts of time.

5 Four Step Approach to Solving Problems (RUBIES) Step 1 RUB Re-Read, Underline the question, Bracket the information and draw a picture. Step 2 I Identify the variables, list the symbols and data, Write down what you know and what are you trying to find Step 4 S Solve the problem, plug-in the numbers and solve. Select the correct answer (if multiple choice) Step 3 E Eliminate unnecessary data and select appropriate formula. Set up the formula.

6 “A car traveled 110 miles in 2 hours.” Consider the problem… d = t = s = FormulaPlug-inAnswer Units, units, units! Step 1 R e-Read, U nderline the question, B racket the information and draw a picture. [110 miles] [2 hours] What is the average speed of the car?

7 “A car traveled 110 miles in 2 hours.” d = t = s = FormulaPlug-inAnswer Units, units, units! 110 miles 2 hours Step 2 I dentify the variables, list the symbols and data, Write down what you know and what are you trying to find [110 miles] [2 hours] What is the average speed of the car?

8 “A car traveled 110 miles in 2 hours.” d = t = s = FormulaPlug-inAnswer Units, units, units! 110 miles 2 hours Step 3 Eliminate unnecessary data and select appropriate formula. Set up the formula. s = d t

9 “A car traveled 110 miles in 2 hours.” d = t = s = FormulaPlug-inAnswer Units, units, units! 110 miles 2 hours Step 3 Set up the formula. EQUATION d t S =

10 “A car traveled 110 miles in 2 hours.” d = t = s = FormulaPlug-inAnswer Units, units, units! 110 mi 2 hours Step 4 Solve the problem, plug-in the numbers and solve. Select the correct answer (if multiple choice) d t 110 mi 2 hr 55mi/hr 55mi/hr 55mi/hr S =

11 Problem Set (speed)

12 “A runner’s average speed during the 10 kilometer race was 20 km/hr. What was his time?’” Consider the problem… d = t = s = FormulaPlug-inAnswer Units, units, units! [10 km] Speed of runner: [20 km/hr] Step 1 R e-Read, U nderline the question, B racket the information and draw a picture.

13 d = t = s = FormulaPlug-inAnswer Units, units, units! 10 km 20 km/hr 10 km Speed of runner: 20 km/hr “A runner’s average speed during the 10 kilometer race was 20 km/hr. What was his time?” Step 2 I dentify the variables, list the symbols and data, Write down what you know and what are you trying to find

14 FormulaPlug-inAnswer Units, units, units! “A runner’s average speed during the 10 kilometer race was 20 km/hr. What was his time?’” d = t = s = 10 km 20 km/hr t = d s t (hr) = d (km) s (km/hr ) Step 3 Eliminate unnecessary data and select appropriate formula. Set up the formula.

15 FormulaPlug-inAnswer Units, units, units! 10 km 20 km/hr d = t = s = 10 km 20 km/hr 0.5 hr “A runner’s average speed during the 10 kilometer race was 20 km/hr. What was his time?’” Time: 0.5 Hour t = d s t (hr) = = 1 r 0.5 hr 10 km 20 km/hr Step 4 Solve the problem, plug-in the numbers and solve. Select the correct answer (if multiple choice)

16 Problem Set (time)

17 “You decide to go to Dallas to see friends. Your friends tell you that it takes 4 hours to get to Dallas at an average speed of 70 miles per hour. Approximately how many miles is it to their house?” Step 1 Re-Read, Underline the question, Bracket the information and draw a picture. Step 2 Identify the variables, list the symbols and data, Write down what you know and what are you trying to find Step 4 Solve the problem, plug-in the numbers and solve. Select the correct answer (if multiple choice) Step 3 Eliminate unnecessary data and select appropriate formula. Set up the formula. FormulaPlug-inAnswer Units, units, units! d = t = s = 280 mi 70 mi/hr 4 hr ? 70 miles per hour d = s * t = 280 mi d =70 mi/hr * 4 hr

18 Problem Set (distance)

19 Instantaneous speed (Reading on your speedometer) and Average speed (Total distance traveled over total time) Both do not involve direction.

20 What is the difference between speed and velocity? 55 mi/hr All of these cars had different velocities because they were traveling in different directions. Velocity has speed & direction.

21A distance/time graph makes it possible to “see” speed. This graph shows how fast the swimmers went during their workout. Which swimmer swam at a constant (the same) speed throughout her workout? Which one stopped during his/her workout? Constant speed Stopped here 400 meters at 10, 15, & 20 minutes Is a straight line A distance/time graph makes it possible to “see” speed. This graph shows how fast the swimmers went during their workout.

22 Make the speed graph & answer some questions

23 Acceleration

24 Acceleration is defined as the change in velocity over time. final velocity – initial velocity final time – initial time acceleration = (change) in velocity (change) time acceleration = V t a = V f - V i t f - t i =

25 A go-cart started from the top of a hill at 5 meters per second. At the bottom of the hill it ended up with a speed of 35 meters per second 6 seconds later. What was the acceleration of the go-cart? Step 1 Re-Read, Underline the question, Bracket the information and draw a picture. top bottom 35 m/s 5 m/s In 6 s V f = Vi=Vi= t = a = FormulaPlug-inAnswer

26 top bottom 35 m/s 5 m/s 6 s V f = Vi=Vi= t = a = FormulaPlug-inAnswer Step 2 Identify the variables, list the symbols and data, Write down what you know and what are you trying to find 35 m/s 5 m/s 6 s Finish: final Velocity Start: initial Velocity final velocity – initial velocity time acceleration = A go-cart started from the top of a hill at 5 meters per second. At the bottom of the hill it ended up with a speed of 35 meters per second 6 seconds later. What was the acceleration of the go-cart?

27 top bottom 35 m/s 5 m/s 6 s V f = Vi=Vi= t = a = 35 m/s 5 m/s 6 s Final Velocity Initial Velocity Step 3 Eliminate unnecessary data and select appropriate formula. Set up the formula. FormulaPlug-in V f - V i t f - t i Answer V f - V i t A go-cart started from the top of a hill at 5 meters per second. At the bottom of the hill it ended up with a speed of 35 meters per second 6 seconds later. What was the acceleration of the go-cart?

28 V f = Vi=Vi= t = a = FormulaPlug-inAnswer 35 m/s 5 m/s 6 s V f - V i t 35m/s – 5 m/s 6s 30 m/s 6 s = 5 m/s 2 Step 4 Solve the problem, plug-in the numbers and solve. Select the correct answer (if multiple choice) 5 m/s 2 A go-cart started from the top of a hill at 5 meters per second. At the bottom of the hill it ended up with a speed of 35 meters per second 6 seconds later. What was the acceleration of the go-cart?

29 (Problem Set)

30 This brings up an important point. In common language, when things speed up, we say that they are "accelerating," and, when they slow down, we say that they are "decelerating." Any time an object's velocity is changing, we say that the object is accelerating. Acceleration is defined as the change in velocity over time.

31 However, in the language of physics, we say that both objects are accelerating, not because both objects are speeding up, but because both objects have changing velocities. POSITIVE (+) ACCELERATION (SPEEDING UP) NEGATIVE (-) ACCELERATION (DECELERATING) SLOW DOWN

32 Velocity involves both speed and direction. Changing velocity does not have to necessarily involve a change in speed. It could just involve a change in direction. 70 mi/h

33 Constant Speed of 55 mph 1. Consider a car moving at a constant speed of 55 mph while turning in a circle. 2. The car's velocity is not constant, even though the speed is constant. 3. WHY? This is because the direction of motion is constantly changing while the car is turning around the track. 4. Since the direction is changing, even though the speed is not, the velocity is changing (velocity involves both speed and direction). Think Differently About Acceleration

34 Constant Speed of 55 mph 6. As a result, the car is accelerating, even though it is neither speeding up nor slowing down. 5. The car is accelerating because its velocity is changing. Think Differently About Acceleration

Obj 12.. Momentum is a property a moving object has because of its mass and velocity. Which would hurt you more if they ran into you?

Obj 12. Momentum is a property a moving object has because of its mass and velocity. Momentum can be calculated by : P = mv. P = momentum (kg  m/s) m = mass (kg) v = velocity (m/s)

Momentum practice problems a. What is the momentum of a 5 kg ball that has a velocity of 15m/s? Given:Formula Subst. Answer & Unit

Momentum practice problems a. What is the momentum of a 5 kg ball that has a velocity of 15m/s? Given:Formula Subst. Answer & Unit P= PP = M V M= 5kg V= 15m/s

Momentum practice problems a. What is the momentum of a 5 kg ball that has a velocity of 15m/s? Given:Formula Subst. Answer & Unit P= PP = M V P = 5 x 15 M= 5kg V= 15m/s

Momentum practice problems a. What is the momentum of a 5 kg ball that has a velocity of 15m/s? Given:Formula Subst. Answer & Unit P= PP = M V P = 5 x kg m/s M= 5kg V= 15m/s

b. What is the mass of a box that has a momentum of 10 kg m/s and a velocity of 5m/s? Given: Formula Subst. Answer & Unit P = M V Momentum Example problems

b. What is the mass of a box that has a momentum of 10 kg m/s and a velocity of 5m/s ? Given: Formula Subst. Answer & Unit P= 10 kg m/s P = M V M= M V= 5m/s Momentum Example problems

b. What is the mass of a box that has a momentum of 10 kg m/s and a velocity of 5m/s ? Given: Formula Subst. Answer & Unit P= 10 kg m/s P = M V 10 = M x 5 M= M V= 5m/s Momentum Example problems

b. What is the mass of a box that has a momentum of 10 kg m/s and a velocity of 5m/s ? Given: Formula Subst. Answer & Unit P= 10 kg m/s P = M V 10 = M x 5 M= M 5 5 V= 5m/s Momentum Example problems

b. What is the mass of a box that has a momentum of 10 kg m/s and a velocity of 5m/s ? Given: Formula Subst. Answer & Unit P= 10 kg m/s P = M V 10 = M x 5 2 kg M= M 5 5 V= 5m/s Momentum Example problems

c. What is the velocity of a child that has a mass of 20kg and a momentum of 50 kg m/s ? Given:Formula Subst. Answer & Unit P = 50 kg m/s P = M V 50 = 20 x V M= 20kg V= V

c. What is the velocity of a child that has a mass of 20kg and a momentum of 50 kg m/s ? Given: Formula Subst. Answer & Unit P = 50 kg m/s P = M V 50 = 20 x V M= 20kg V= V

c. What is the velocity of a child that has a mass of 20kg and a momentum of 50 kg m/s ? Given: Formula Subst. Answer & Unit P= 50 kg m/s P = M V 50 = 20 x V 5/2 or 2.5 M= 20kg m/s V= V Top first …. So 50 ÷ 20

Copy this question and the steps to solve! d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P = M V Our first job is to find the momentum of the Miata…………

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V M=900 kg V= 30 m/s Our first job is to find the momentum of the Miata…………

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 M=900 kg V= 30 m/s

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Since they did NOT move after the collision, their “momentums” must have been EQUAL!

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Since they did NOT move after the collision, their “momentums” must have been EQUAL! Given/Civic: P = 27,000 kg m/s

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Since they did NOT move after the collision, their “momentums” must have been EQUAL! Given/Civic: P = 27,000 kg m/s M = 1200kg V = V

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Since they did NOT move after the collision, their “momentums” must have been EQUAL! Given/Civic: P = 27,000 kg m/s 27,000 = 1200 x V M = 1200kg P = MV V = V

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Given/Civic: P = 27,000 kg m/s 27,000 = 1200 x V M = 1200kg P = MV V = V Since they did NOT move after the collision, their “momentums” must have been EQUAL!

d. If a Miata (900kg) at 30 m/s runs head-on into a Civic (1200kg) and neither car moves after the collision, what is the velocity of the Civic ? Given/ Miata: Miata P= P P = M V P = 900 x 30 P=27,000 M=900 kg kg m/s V= 30 m/s Since they did NOT move after the collision, their “momentums” must have been EQUAL! Given/Civic: P = 27,000 kg m/s 27,000 = 1200 x V M = 1200kg P = MV V = V V = 22.5 m/s That was the Velocity of the Civic!

+ Obj 13. The Law of Conservation of Momentum - the total amount of momentum of a group of objects does not change unless outside forces act on the objects. (Ex – momentum of cue ball = momentum of balls struck + remaining momentum in cue ball) What outside force acts against the balls now?

If a larger truck were to run, head to head, into a smaller truck moving at the same speed, what would happen to the momentum of each truck?