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Presented by: Ms. Bowie.  By the end of the Presentation you should: ◦ Understand the difference between position, distance and displacement; ◦ Understand.

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Presentation on theme: "Presented by: Ms. Bowie.  By the end of the Presentation you should: ◦ Understand the difference between position, distance and displacement; ◦ Understand."— Presentation transcript:

1 Presented by: Ms. Bowie

2

3  By the end of the Presentation you should: ◦ Understand the difference between position, distance and displacement; ◦ Understand the difference between speed and velocity; ◦ Be able to solve simple problems involving average speed and velocity; ◦ Carry out simple calculations involving distance, displacement, speed, velocity and acceleration.

4  We can describe an object’s position, in one dimension, using a number line.  The initial position can be recorded as “zero” on the number line.  In this case, my crazy frog ended up at position “9” on the number line.

5  Position can also include information about where the object ends up relative to the initial position using + and – signs.  Again, in this case, my crazy frog ended up at position +9.

6  By convention, a positive integer generally represents motion to the right or up.  A negative integer generally represents motion to the left or down.  The symbol for position is usually “x”.

7  For example, assuming divisions are 1 m, we could say: ◦ The rocket is +3 m relative to the ghost. ◦ The teddy driving car is +6 m relative to the ghost. ◦ The rocket is -3 m relative to the teddy driving car. ◦ The ghost is -6 m relative to the teddy driving car.

8  Distance is how much an object’s position changes.  How far an object has travelled is referred to as the total distance travelled (or simply, it’s distance).  The unit of measure for distance is the meter (m).  Distance is a scalar.  Scalar means that it has magnitude, but no direction is given.

9  Practice Problem #1  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. What distance did the elephant travel? 500 m300 m 1300 m

10  Practice Problem #1 - SOLUTION  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. What distance did the elephant travel?  x = d 1 + d 2 + d 3  ∴x = 1300 m + 500 m + 300 m = 2100 m  Notice that there is no direction noted as distance is a scalar. d 2 = 500 md 3 = 300 m d 1 =1300 m

11  Displacement (Δd) is a vector that describes a straight line from the starting position to the final position.  Displacement (d f – d i ) is also measured in meters (m)  It is sometimes called the position vector as it points from your initial position to the final vector.

12  Problem #2  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. What is the displacement? d 2 = 500 md 3 = 300 m d 1 =1300 m Displacement vector didi dfdf

13  Problem #2  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. What is the displacement?  Δd = d f – d i = 500 m east d 2 = 500 md 3 = 300 m d 1 =1300 m Displacement vector didi dfdf 500 m

14  Average speed is the rate at which distance is travelled. It is a scalar.  It is measured in m/s  Speed is scalar (to help you remember) Note: the line above indicates “average”

15  Problem #3  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. What is the average speed if the entire journey took 600 seconds?  V = 2100 m = 3.5 m/s 600 s Since speed is a scalar we do not need to indicate direction.

16  Average velocity is the rate at which displacement changes. It is a vector.  It is measured in m/s  Velocity is Vector (to help you remember) Note: the line above indicates “average”

17  Problem #3  An elephant walked 1300 m East to the watering hole. It then walked 500 m West to its favourite food, the acacia tree. However, it was scared by a pack of hyenas and ran 300 m West. The entire journey took 600 s. What is the average velocity?  v =Δx/Δt = d f – d i / t= 500 m – 0 m /600 s  v =500 m / 600 s  v =0.833 m/s east dfdf d 2 = 500 md 3 = 300 m d 1 =1300 m Displacement vector didi 500 m

18  Sid travels 4 m east and 3 m north in search of an acorn. The entire journey takes him 20 seconds. Find: 1.Sid’s distance travelled 2.Sid’s displacement 3.Sid’s average speed 4.Sid’s average velocity

19  Sid travels 4 m east and 3 m north in search of an acorn. The entire journey takes him 20 seconds. 1.Sid’s distance travelled Δd = d 1 + d 2 Δd = 4 m + 3 m = 7 m (Notice that no direction is given because distance is a scalar) 4 m E 3 m N

20  Sid travels 4 m east and 3 m north in search of an acorn. The entire journey takes him 20 seconds. 1.Sid’s displacement Δd = d f - d i c 2 = a 2 + b 2 c 2 = 4 2 + 3 2 c 2 = 16+ 9 c = √25 c = 5 m 4 m E 3 m N Δd = df – d i didi dfdf

21  Sid travels 4 m east and 3 m north in search of an acorn. The entire journey takes him 20 seconds. 1.Sid’s average speed v = 7 m/ 20 s v = 0.35 m/s 4 m E 3 m N

22  Sid travels 4 m east and 3 m north in search of an acorn. The entire journey takes him 20 seconds. 1.Sid’s displacement v = 5m/20 s c = 0.25 m/s NE 4 m E 3 m N Δd = df – d i didi dfdf

23  Acceleration is the rate at which velocity changes.  Acceleration is a vector.  The units are m/s/s or simply m/s 2

24  The units m/s 2 might sound odd but think of it like this:  If you have an acceleration of 5 m/s 2 this means that your speed is increasing by 5 m/s every second. So: ◦ at time 1 the speed is 5 m/s. ◦ at time 2 s, the speed is 10 m/s and ◦ at time 3 s, the speed is 15 m/s…

25 Problem #4 Marty the Moose accelerates from rest to a velocity of 9 m/s in a time span of 3 seconds northward on Cole Harbour Rd. Calculate Marty’s acceleration. a = v f – v i t f – t i a = 9 m/s – 0 m/s = 9 = 3 m/s 2 N 3 – 0 3

26 d f (m) d i (m) Δd (m) t f (s) t i (s) Δt (s) Velocity (m/s) +20.0+17.36.55.0 +26.9+5.38.32.1 +26.8-15.40105.4 55012510 60-20010030-2 Complete the Chart http://www.youtube.com/watch?v=rZo8-ihCA9E&feature=related


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