1. Linear Kinematics Chapter 2 in the text 5/2/2019
Dr. Sasho MacKenzie - HK 376

2. KINEMATICS LINEAR ANGULAR Future Class
Scalars
Distance
Speed
Future Class
Vectors
Displacement
Velocity
Acceleration
5/2/2019
Dr. Sasho MacKenzie - HK 376

3. Scalars A measure that only considers magnitude
Does not consider direction
E.g., a length of 2.7 meters is a scalar measure
2.7 m
The javelin is 2.7 meters but has no direction
5/2/2019
Dr. Sasho MacKenzie - HK 376

4. Vectors Describes both a magnitude and direction
E.g., a displacement of 15 meters in the positive direction is a vector.
Represented by arrows, in which the length represents magnitude and orientation represents direction.
The javelin meters in the positive direction
5/2/2019
Dr. Sasho MacKenzie - HK 376

5. Distance
A measure of the length of the path followed by an object from its initial to final position.
A scalar quantity (no direction)
5/2/2019
Dr. Sasho MacKenzie - HK 376

6. Speed The rate of motion of an object
The rate at which an object’s position is changing.
A scalar quantity (no direction)
5/2/2019
Dr. Sasho MacKenzie - HK 376

7. Displacement
The straight-line distance in a specific direction from the starting position to the ending position.
A vector quantity (must have direction)
As the crow flies
5/2/2019
Dr. Sasho MacKenzie - HK 376

8. Velocity The rate of motion in a specific direction
Similar to speed but with a direction
A vector quantity
5/2/2019
Dr. Sasho MacKenzie - HK 376

9. Distance vs. DisplacementW E S
Distance vs. Displacement
End
10 km
Start
5 km
11.2 km
63.4°
Distance
5 km + 10 km = 15 km
Displacement
Magnitude:  (5 km)2 + (10 km)2 = 11.2 km
Direction: Inverse Tan (10 km / 5 km) = 63.4° East of North
5/2/2019
Dr. Sasho MacKenzie - HK 376

10. Speed vs. Velocity Speed Velocity
It took Bill 3.5 hours in total to walk 5 km North and 10 km East. What was Bill’s average speed and average velocity?
Speed
= Distance/Time
= 15 km/ 3.5 hours
= 4.3 km/h
Velocity
= Displacement/Time
= 11.2 km/ 3.5 hours
= 3.2 km/h 63.4° E of N
5/2/2019
Dr. Sasho MacKenzie - HK 376

11. Acceleration The rate at which an object’s speed or velocity changes.
When an object speeds up, slows down, starts, stops, or changes direction, it is accelerating.
Always a vector quantity (has direction)
5/2/2019
Dr. Sasho MacKenzie - HK 376

12. Acceleration
The direction of motion does not indicate the direction of acceleration.
An object can be accelerating even if its speed remains unchanged. The acceleration could be due to a change in direction not magnitude.
5/2/2019
Dr. Sasho MacKenzie - HK 376

13. Circle Circumference = 2r; Circle Diameter = 2r; r is radius
Sample Problem
Bolt runs 200 m in seconds. Assume he ran on the inside line of lane 1, which makes a semicircle (r = 36.5 m) for the first part of the race. He runs the curve in 11 s.
Circle Circumference = 2r; Circle Diameter = 2r; r is radius
What distance was run on the curve?
What was his displacement after the curve?
Total distance?
Total displacement?
Average velocity on the curve?
Average speed on the curve?
Average velocity for the race?
Average speed for the race?
Start N W E S
36.5 m
Finish
5/2/2019
Dr. Sasho MacKenzie - HK 376

14. Instantaneous Velocity
The average velocity over an infinitely small time period.
Determined using calculus
The derivative of displacement
The slope of the displacement curve
5/2/2019
Dr. Sasho MacKenzie - HK 376

15. Instantaneous Acceleration
The average acceleration over an infinitely small time period.
Determined using Calculus
The derivative of velocity
The slope of the velocity curve
5/2/2019
Dr. Sasho MacKenzie - HK 376

16. Slope Y (4,8) 8 (0,0) X 4 Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2
Slope could be anything not just velocity: give other example 4
Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2
run X X2 – X –
5/2/2019
Dr. Sasho MacKenzie - HK 376

17. Velocity is the slope of Displacement
(4,8) 8
Displacement
(m)
(0,0) X 4
Time (s)
Average
Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
run t t2 – t – s
5/2/2019
Dr. Sasho MacKenzie - HK 376

18. The displacement graph on the previous slide was a straight line, therefore it’s slope was 2 at every instant.
Which means the velocity at any instant is equal to the average velocity.
However if the graph was not straight the instantaneous velocity could not be determined from the average velocity.
5/2/2019
Dr. Sasho MacKenzie - HK 376

19. Average vs. InstantaneousX Y 8 4
Displacement
(m)
Time (s)
The average velocity does not accurately represent slope at this particular point.
(0,0)
(4,8)
Average
Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
run t t2 – t – s
Read Ch. 6 pages for next class
5/2/2019
Dr. Sasho MacKenzie - HK 376

20. Circle Circumference = 2r; Circle Diameter = 2r; r is radius
Sample Problem
Bolt runs 200 m in seconds. Assume he ran on the inside line of lane 1, which makes a semicircle (r = 36.5 m) for the first part of the race. He runs the curve in 11 s.
Circle Circumference = 2r; Circle Diameter = 2r; r is radius
What distance was run on the curve?
What was his displacement after the curve?
Total distance?
Total displacement?
Average velocity on the curve?
Average speed on the curve?
Average velocity for the race?
Average speed for the race?
114.7 m
73 m South
Start
200 m
112.3 m 40.5° South of East N W E S
6.64 m/s South
36.5 m
m/s
5.85 m/s 40.5° South of East
m/s
Finish
5/2/2019
Dr. Sasho MacKenzie - HK 376