Kinematics Lesson Objectives At the end of lesson

Kinematics

Lesson Objectives At the end of lesson, students should be able to: § § § . State what is meant by distance and displacement. State what is meant by speed and velocity, and calculate average speed and velocity. State what is meant by uniform acceleration, and calculate acceleration.

House You are visiting a friend on his island. One day, he shows you a map of the island. At the bottom of the map there is this message. “Dig 3 km away from the house for the treasure. ” Both of you decide to set off to find this treasure.

Treasure Map Path J: 3 km Path D: 1. 5 km Path A: 2 km Path I: 2 km Path E: 3 km Path C: 3 km Path B: 1 km Path H: 1 km Path G: 3 km Path F: 1. 5 km

Treasure Map l Where do you start to dig for the treasure? l Is there enough information to know where to dig? l What other information do we need?

Recall l l What is a vector? Let us look at this table below and fill in the blanks. Physical Quantities Scalars or Vectors Length Scalar Mass Scalar Weight Vector Temperature Scalar Moment Vector

Distance and Displacement l l Distance § A scalar quantity § Magnitude only. Displacement § A vector quantity § Magnitude and direction

Treasure Map Path J: 3 km Path D: 1. 5 km Path A: 2 km Path I: 2 km Path E: 3 km Path C: 3 km Path B: 1 km Path H: 1 km Path G: 3 km Path F: 1. 5 km

Where to dig? l We decide to dig at the hills for the treasure. We walk along path A and D from the house. The total distance we have traveled is 1. 5 km + 1. 5 km = 3 km.

Where to dig? l l However, our displacement is not 3 km. Our displacement is shown by the direct line (in red) Path D: 1. 5 km Displacement Calculation Path A: 1. 5 km Let displacement be s s = 2. 12 km By Pythagoras Theorem s 2 = 1. 52 + 1. 52 450 s = (1. 52 + 1. 52) s = 2. 12 km

Where to dig? Instead of walking to the hills, we decide to dig at the rock. We start from the house and then walk west for 3 km to reach the rock and we dig there. l What is our distance traveled? Distance traveled = 3 km l l What is our displacement? Displacement = 3 km , west

Distance and displacement Question 1 Starting from the house, We walk along path B, path G and path H. What is our (i) Distance traveled. (ii) Displacement. (i) Distance traveled = 1+3+1 = 5 km (ii) Displacement = 3 km, West

Distance and displacement Question 2 Starting from the house, We walk along path C, path I and path J. What is our (i) Distance traveled. (ii) Displacement. (i) Distance traveled = 3+2+3 = 8 km (ii) Displacement = 2 km, north

Distance and displacement Question 3 Starting from the house, We walk along path B. What is our (i) Distance traveled. (ii) Displacement. (i) Distance traveled = 1 km (ii) Displacement = 1 km, south

Distance and displacement Question 4 Starting from the house, We walk along path A, path D, path E, Path F, and path B. What is our (i) Distance traveled. (ii) Displacement. (i) Distance traveled = 2+1. 5+3+1. 5+1 = 9 km (ii) Displacement = 0 Distance vs. Displacement applet

Practice Question 1 l The diagram below shows position of a cross-country skier at various times. The skier moves from A to B to C to D. Determine the distance traveled by the skier and the resulting displacement during these three minutes. Distance Traveled = 180+140+100 = 420 m Total Displacement = 140 m to the right

Practice Question 2 l Diagram below shows several of Beckham’s positions at various times of a football game. Beckham moves from position A to B to C to D. What is Beckham's resulting displacement and distance traveled? Distance Traveled = 35+20+40 = 95 m Total Displacement = 55 m to the left

Speed & Velocity Speed Definition: Speed is defined as the rate of change of distance with time. Equation: v = s / t where v = speed; s = distance; t = time interval

Speed & Velocity Example: A motorcycle covers a distance of 15 km in 30 minutes. Q. What is the motorcycle’s average speed? <v> = s / t = 15 km / (30/60) h = 15 km / 0. 5 h = 30 km h-1 Q. Does the average speed help us to determine the maximum or minimum speed of the motorcycle? No, it does not. Q. The above answer seems slow for a motorcycle. Explain why it is probably correct. Average speed is just an average value for the car. For example, on an Expressway, the motorcycle could be travelling at 80 km/h. After exiting from Expressway, the motorcycle gets stuck in a traffic jam and travels at 5 km/h. As such, it could be travelling faster at times and slower at other times which is why the average speed seems correct.

Speed & Velocity Question 1: A plane travels at a constant speed and covers a distance of 2750 km in a period of 2 hours 30 minutes. What is the speed of the aeroplane? s = 2750 km t = 2 h 30 min = 2. 5 h <v> = s / t = 2750 km / 2. 5 h = 1100 km/h

Speed & Velocity Question 2: A man walks along his garden path in 2 mins. If his average speed for the walk was 2 m/s, how long is his garden path? t = 2 mins = 120 s <v> = 2 m/s s = <v> x t = (2 x 120) m = 240 m Note the SI units of speed and time in this example.

Speed & Velocity Question 3: A man jogs a distance of 8 km in 40 minutes. Calculate his average speed in (i) km/h & (ii) m/s. (i) <v> = s / t = 8 km / (40/60)h = 12 km/h (ii) <v> = s / t = (8 x 1000)m / (40 x 60)s = 3. 33 m/s

Speed & Velocity Question 4: Convert the following speeds between m/s and km/h. i) 15 m/s & ii) 100 km/h (i) v = 15 m/s = (15/1000)km / (1/60 x 60)h = 54 km/h (ii) v = 100 km/h = (100 x 1000)m / (1 x 3600)s = 27. 8 m/s

Speed & Velocity Question 5: You walk a distance of 2 km at an average speed of 8 km/h. You then run for a further 5 minutes with an average speed of 14 km/h. What is your average velocity for the whole journey? s 1 = 2 km v 1 = 8 km/h t 1 = s 1 / v 1 = 2/8 h = ¼ h v 2 = 14 km/h t 2 = 5 min = 5/12 h s 2 = v 2 x t 2 = 14 km/h x 5/12 h = 35/6 km <v> = (s 1 + s 2) / (t 1 + t 2) = (35/6 + 2)km / (1/4 + 5/12)h <v> = 11. 75 km/h

Speed & Velocity is a vector and has both magnitude and direction. . Definition: Velocity is defined as the rate of change of distance in a specified direction with time. Velocity can also be defined as the rate of change of displacement with time.

Speed & Velocity Example: A car drives 40 km north along a road and then 20 km south along the same road. It takes 2 hours to complete its journey. Calculate: a) average speed for the journey ; b) the average velocity for the journey. a) Total Distance = 40 + 20 km = 60 km Time = 2 h Average speed = distance / time = 60 km / 2 h = 30 km/h b) Displacement = (40 – 20) km = 20 km Time = 2 h Average velocity = displacement / time = 20/2 = 10 km/h (towards north).

Acceleration is defined as the rate of change of velocity with respect to time. Equation: a = (v – u) t where a = acceleration in m s-2 ; v = final velocity in m s-1 ; u = initial velocity in m s-1 ; t = time interval between the change from the initial velocity to the final velocity.

Acceleration Example 1: A bus at rest is accelerated to a velocity of 30 m/s in 10 seconds. Calculate the acceleration of the bus. Solution: Initial velocity, u = 0 m s-1. Final velocity = v = 30 m s-1. Time interval = t = 10 s. a = (v – u) / t = [(30 – 0) / 10] m s-2 = 3 m s-2.

Acceleration Question 1: A car is travelling at a constant velocity of 20 m/s before it accelerates to a velocity of 65 m/s in 5 seconds. What is the acceleration of the car? Solution: u = 20 m s-1. v = 65 m s-1. t = 5 s. a = (v – u) / t = [(65 – 20) / 5] m s-2 = 9 m s-2.

Acceleration Question 2: A car initially has a velocity of 60 m/s. After braking for 5 seconds it has a velocity of 20 m/s. Calculate the acceleration of the car. Solution: u = 60 m s-1. v = 20 m s-1. t = 5 s. a = (v – u) / t = [(20 – 60) / 5] m s-2 = - 8 m s-2. Note the negative sign for the acceleration in this case.

Acceleration Deceleration When an object reduces its velocity (or speed), we can state this in several way. We can say that the object (a) is experiencing deceleration ; (b) is experiencing retardation ; (c) is experiencing negative acceleration.

Acceleration Question 3: A train slows down from 80 m/s to rest in one minute. Calculate the retardation of the train. Solution: u = 80 m/s ; v = 0 m/s ; t = 1 min = 60 s ; a = (v-u) / t = [(0 – 80) / 60] m/s 2 = - 1. 33 m/s 2 Since acceleration is – 1. 33 m/s 2 , retardation is 1 m/s 2.

Concept Map All these quantities are Displacement Is the rate of change of displacement per unit time Velocity Vectors. v =s t They have both magnitude and Is the rate of change of velocity per unit time Acceleration direction. a = (v – u) Write down one thing you have learnt in this lesson. t