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Genazzano FCJ College Year 12 Motion Motion along a straight line

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Presentation on theme: "Genazzano FCJ College Year 12 Motion Motion along a straight line"β€” Presentation transcript:

1 Genazzano FCJ College Year 12 Motion Motion along a straight line

2 Person outside the bus 5 10 The bus moved away from the tree
5 10 The bus moved away from the tree The person is comparing the position of the bus with respect to the position of the tree Reference (or origin) is position of the tree

3 Person inside the bus 10 5 The tree moved away from the bus.
The tree moved away from the bus. The person is comparing the position of the tree with respect to the position of the bus. Reference (or origin) is position of the bus.

4 Motion is relative Both the observations are correct. The difference is what is taken as the origin. Motion is always relative. When one says that a object is moving, he/she is comparing the position of that object with another object. Motion is therefore change in position of an object with respect to another object over time. Kinematics studies motion without delving into what caused the motion.

5 Q. How much distance do you have to travel to reach school?
Direct Path (1.1 km) Actual Path (2 km) Q. How much distance do you have to travel to reach school? Q. If you were to draw a straight line between your house and school, what would be the length of that line?

6 Q. How much distance do you travel in one round trip to the school?
Direct Path (1.1 km) Actual Path (2 km) Q. How much distance do you travel in one round trip to the school? Q. After one trip how far away are you from your home?

7 Distance and Displacement
Distance = length of the actual path taken to go from source to destination Displacement = length of the straight line joining the source to the destination or in other words the length of the shortest path

8 Checkpoint Suppose it was given that the person started by point A and walked in a straight line for 5 km. Can you calculate the end point of his/her journey? No, the person could be anywhere on the circle of 5 km radius. A Unless we know the direction of the motion we cannot calculate the end point of the journey.

9 Sample Problem Robert and Sarah both start from their house. Robert walks 2 km to the east while Sarah walks 1 km to the west and then turns back and walks 1 km. Distance travelled by them is the same (2 km) Is their displacement also the same? No – Sarah is back home and her displacement is 0 m. This is because direction of motion is different in both cases. You require both distance and direction to determine displacement.

10 Distance travelled = 7 km, Displacement = 5 km from A towards C.
Sample Problem C Distance AB = 3 km due East Distance BC = 4 km due North What is the distance travelled by a person who moves from A to C via B? What is the displacement? What is the direction of the displacement? A B Distance travelled = 7 km, Displacement = 5 km from A towards C.

11 Rate of Motion Distance travelled per unit time or the displacement per unit time. 𝑠𝑝𝑒𝑒𝑑= π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘‘π‘–π‘šπ‘’ π‘šπ‘’π‘‘π‘Ÿπ‘’π‘ /π‘ π‘’π‘π‘œπ‘›π‘‘ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦= π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘‘π‘–π‘šπ‘’ π‘šπ‘’π‘‘π‘’π‘Ÿ/π‘ π‘’π‘π‘œπ‘›π‘‘ When an object is travelling along a straight line its velocity is equal to its speed.

12 Year 11 Knowledge 1. Kinematics (Notes from booklet) HOMEWORK
Kinematics Worksheet HOMEWORK Text: Chapter 1 Summary

13 Demonstration: Position – time graph using a data logger
Written up in log-book

14 2. Forces Newton’s 3 Laws Notes 1643 - 1727 Week 1: Worksheet 1
Due Date:

15 3. Mass & Weight Mass & Weight Notes
How do theme park rides and elevators change how we feel? Mass & Weight Notes Elevator Notes from Chapman

16 3a. Acceleration of Gravity
Objects that fall to the Earth all experience an acceleration. The acceleration due to gravity is g = 9.8 m/s2. In our tests and examinations this will be 10. This acceleration must be due to a force.

17 3b. Force of Gravity The acceleration of a falling mass m is -g.
The force on the mass is found from F = ma (action). This gravitational force is F = -mg. Kinematic view Dynamic view

18 3c. Normal Weight We measure weight with a scale that measures normal force. W = mg Weight is related to mass by the gravitational field g.

19 3d. False Weight A vertical acceleration can change the weight.
The normal force on the floor is our sense of weight. Downward acceleration reduces weight Upward acceleration increases weight Mass is unchanged. Newton’s law of acceleration F = ma net force, F = -mg + FN. Solve for the normal force -mg + FN = ma FN = ma + mg FN = m (a + g) Apparent mass based on g mapp = FN / g

20 3e. Accelerated Weight An elevator is accelerating downward at 2.0 m/s2. The person has a mass of 70 kg. What mass is on the scale? Add all the forces, but the net force is – ma = FN – mg. Solve for FN = m (g – a) Convert to mass mapp = FN /g The scale shows 56 kg.

21 3f. Weightlessness If the elevator accelerated downward at g, the normal force would become 0. FN = m (g – a) = m (g – g) = 0 The person would feel weightless. An object in free fall is weightless, but not massless.

22 4. Projectile Motion Projectile Motion Simulation A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

23 It can be understood by analyzing the horizontal and vertical motions separately.

24 The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly. Projectile Motion Notes

25 Sample Problem r = 100 m The adjoining figure shows a Formula 1 racing track. A driver is did 10 laps, what is the distance travelled by the driver at the end of the race? What is the displacement? If the driver took seconds to complete the laps, what is his speed and velocity in km/hr? Distance = 2βˆ—Ξ βˆ—π‘Ÿ βˆ—π‘™π‘Žπ‘π‘  = 6280 m, Displacement = 0 m Speed = π‘‘π‘–π‘ π‘Žπ‘›π‘π‘’ π‘‘π‘–π‘šπ‘’ π‘š/𝑠 = π‘š/𝑠 = 180 km/hr

26

27 Uniform Motion 25 20 15 10 5 30 40 50 Distance (m) Time (s) Distance – Time graph A distance – time graph represents the distance travelled with respect to time. When an object covers equal distance in every time interval, it is said to be having uniform motion. In an uniform motion, the speed of the object remains constant. A stationary body is also an example of uniform motion

28 Velocity – Time graph Distance travelled = 20 m
25 20 15 10 5 30 40 50 Distance (m) Time (s) Distance – Time graph 1.25 1.0 0.75 0.5 0.25 10 20 30 40 50 Speed (m)/s Time (s) Speed – Time graph Distance travelled = 20 m Area of shaded region = 0.5 * 40 = 20m

29 Uniform and Non-Uniform Motion
1.25 1.0 0.75 0.5 0.25 10 20 30 40 50 Velocity (m)/s Time (s) Velocity – Time graph Uniform Motion Acceleration = 0 m/s2 1.25 1.0 0.75 0.5 0.25 10 20 30 40 50 Velocity (m)/s Time (s) Velocity – Time graph Non-uniform Motion Acceleration = m/s2

30 Rate of Change of Velocity
acceleration = π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ π‘‘π‘–π‘šπ‘’ meter/second2 A body is said to be accelerating if there is a change in velocity. Velocity has magnitude and direction. A body has acceleration when either of them changes.

31 Uniform and Non-Uniform Acceleration
1.25 1.0 0.75 0.5 0.25 10 20 30 40 50 Velocity (m)/s Time (s) Velocity – Time graph Uniform Acceleration Acceleration = m/s2 1.25 1.0 0.75 0.5 0.25 10 20 30 40 50 Velocity (m)/s Time (s) Velocity – Time graph Non-uniform Acceleration

32 Sample Problem Which object has the maximum acceleration?
25 20 15 10 5 30 40 50 Velocity (m/s) Time (s) A B C D Which object has the maximum acceleration? Which object has no acceleration? How much distance is covered by object D in 20 seconds? Explain the motion represented by D. Given an example of such a motion in real life.

33 1st Equation of Motion Initial velocity = u Final velocity = v
t Velocity (m)/s Time (s) Velocity – Time graph Uniform Acceleration Initial velocity = u Final velocity = v Time = t Acceleration = a Displacement = s Acceleration = Rate of change of velocity π‘Ž= 𝑣 βˆ’π‘’ 𝑑 π‘š/ 𝑠 2 π‘Žπ‘‘=𝑣 βˆ’π‘’ 𝒗=𝒖+𝒂𝒕 π’Ž/𝒔

34 2nd Equation of Motion Initial velocity = u Final velocity = v
t Velocity (m)/s Time (s) Velocity – Time graph Uniform Acceleration Initial velocity = u Final velocity = v Time = t Acceleration = a Displacement = s Displacement = Area under the line 𝑠=π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’+π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑠=𝑒𝑑 𝑑 βˆ— 𝑣 βˆ’π‘’ π‘š But (π‘£βˆ’π‘’) 𝑑 =π‘Ž or 𝑣 βˆ’π‘’ =π‘Žπ‘‘ 𝒔=𝒖𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐 π’Ž

35 3rd Equation of Motion Initial velocity = u Final velocity = v
t Velocity (m)/s Time (s) Velocity – Time graph Uniform Acceleration Initial velocity = u Final velocity = v Time = t Acceleration = a Displacement = s Displacement = Area under the line 𝑠=π‘Žπ‘Ÿπ‘’π‘Ž π‘‘π‘Ÿπ‘Žπ‘π‘’π‘§π‘–π‘’π‘š 𝑠= 𝑒+𝑣 βˆ—π‘‘ π‘š But (π‘£βˆ’π‘’) π‘Ž =t 𝑠= 𝑒+𝑣 βˆ— (π‘£βˆ’π‘’) π‘Ž π‘š ∴2π‘Žπ‘ = 𝑒+𝑣 βˆ— 𝑒 βˆ’π‘£ ∴2π‘Žπ‘ = 𝑒 2 βˆ’ 𝑣 2 𝒗 𝟐 = 𝒖 𝟐 +πŸπ’‚π’”

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