1. Days 1 - 2
UNIT 1 Motion Graphs x t
Lyzinski Physics

2. The purpose of this chapter is to learn the 1st step of Mechanics (the study of motion) which is KINEMATICS (the study of motion with no regards to what is causing the motion). The study of what is “causing” the motion is known as dynamics, and we will study this in a later chapter.

3. PHYSICS - Mechanics - Electricity - Magnetism - Optics - Waves
KINEMATICS
- Mechanics
A “description” of motion
- Electricity
DYNAMICS
- Magnetism
A study of what “causes”
motion
- Optics
- Waves

4. Day #1 * Distance * Speed * Scalars * d-t graphs

5. Definition Distance (d) – the length of the path followed by an object
* If an object’s path is straight, the distance is the length of
the straight line between start and finish.
** If an object’s path is NOT straight, the distance is the
length of the path if you were to “straighten it out” and
measure it the way you would measure the length of a
curved shoelace.
start
finish
start
finish

6. B C A
meters
Using the number line above, what would be the distance travelled if an object travelled from ….. 1m
- A to B
- A to C
- A to C and then back to A
- C to B, passing through A 4m
4m + 4m = 8m
4m + 1m = 5m

7. 4 yd B 12 yd 5 yd C A Sally and Timmy are at point A.
Sally walks directly to point C (taking the shortest path).
Timmy also takes a “shortest path”, but has to stop at point B for lunch first.
How much further has Timmy walked when he arrives?
4 yd

8. Definition
Average Speed (s) – the distance travelled during a time interval divided by the elapsed time.
s = d/Dt
(or s=d/t)
Since Dt = t2 – t1, if t1 = 0, then
Dt = t2 – 0 = t2 = t

9. 1 h, 10 min d = 3mi + 4mi = 7mi s = 6 mi/hB C A
miles
Larry walks from point B to point C, and then goes directly to point A. If he walks at an average speed of 6 mph, how long does the trip take him?
d = 3mi + 4mi = 7mi
s = 6 mi/h
s = d/t  t = d/s = (7mi)/(6mi/h)=1.17h
1 h, 10 min
Use appropriate units

10. B C A km
Larry runs from point A to point B in 5 minutes and then proceeds to jog directly to point C, taking his time in 30 additional minutes. Find…
Larry’s average speed during the first portion of the trip.
The average speed during the second portion of the trip.
Larry’s average speed for the entire trip.
s = d/t = (1km)/(5min) = 0.2 km/min = 12 km/h
s = d/t = (3km)/(30min) = 0.1 km/min = 6 km/h
s = d/t = (4km)/(35min) = km/min = 6.86 km/h

11. Definition distance speed time
Scalar – a quantity that has a magnitude only, no direction.
* YES, scalars can have units.
** What scalars have we learned about thus far?
___________ ____________ ___________
distance
speed
time
I thought time could march backward?

12. d-t graphs E F D C B A Constant speed Speeding UP
Constant Speed (faster!)
Slowing Down
At rest
t (sec)
d (m) B C E A D F
120
100 50 30

13. sAB = rise/run = (30-0m) / (10-0s) = 3 m/s
t (sec)
d (m) B C E A D F
120
100 50 30
SLOPE
Speed on a d-t graph can be found by taking the _______________.
sAB = rise/run = (30-0m) / (10-0s) = 3 m/s
sCD = rise/run = (100-50m) / (20-15s) = 10 m/s

14. Open to in your Unit 1 packet 1
d-t graphs CANNOT have sharp points
NOTHING CAN STOP INSTANTANEOUSLY!!
520 – 170yd = 350 yd (approximately)

15. Day #2 * Position * Displacement * Average Velocity * Vectors * x-t graphs

16. Definition
Position (x) – the location of an object with respect to a specified reference point.
*We choose this reference point to be the origin of a
coordinate system. A km
The position of particle “A” is either x = -3 or x = 6, depending on which reference point (or origin) you use.

17. These are all VECTORS. What’s a vector?
Definition
Displacement (Dx) – the change in an object’s position during a time interval.
Dx = x2 – x1 or
Dx = xf – xi
*Displacement must have both a magnitude (size) and a
direction (right, left, up, down, north, south, etc).
These are all VECTORS. What’s a vector?

18. 1m, 1m [right] 4m, 4m [left] 8m, 0m 4m, 3m [left] -3 -2 -1 0 1 B C A A
meters
Using the number line above, find the distance travelled and the displacement in moving from
1m, 1m [right]
- A to B
- C to A
- A to C and then back to A
- C to B, passing through A
Dx = 1 – (1m) = 0m
4m, 4m [left]
8m, 0m
4m, 3m [left]
Dx = (-2) – (1m) = -3m OR 3m [left]

19. (or v=Dx/t) Definition
Average Velocity ( v ) – the displacement of an object divided by the elapsed time.
v = Dx/Dt
(or v=Dx/t)

20. Find Sam’s avg. speed and avg. velocity for the entire trip.B C
Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speed during each interval are as follows.
AB: 7 m/s
BC: 8 m/s
CD: 6 m/s
DA: 7.5 m/s
Find Sam’s avg. speed and avg. velocity for the entire trip.
s = d/t  t = d/s = 100m/7sec = sec
100m/8sec = 12.5 sec
100m/6sec = sec
100m/7.5sec = sec
s = d/t = (400m)/(56.786s) = 7.04 m/sec
Avg Velocity = 0 since Dx = 0 for the entire trip.

21. s = d/t  t = d/s = 200m/(14.286+12.5s) = 7.47 m/s A D B C
Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speeds during each interval are as follows.
AB: 7 m/s, sec
BC: 8 m/s, 12.5 sec
CD: 6 m/s, sec
DA: 7.5 m/s, sec
100
104.94 m
Find Sam’s average speed and average velocity for the 1st half of the race.
s = d/t  t = d/s = 200m/( s) = 7.47 m/s
v = Dx/t = (104.94m )/( s) = 3.92 m/sec

22. Definition position velocity
Vector – a quantity that has both magnitude AND a direction … oh yeh!
* YES, vectors can have units.
** What vectors have we learned about thus far?
____________ ________________ ___________
position
displacement
velocity

23. Scalars vs. Vectors Displacement:
has magnitude & direction (example: 15 cm east)
Distance:
has a magnitude only (example: 6 ft) 1 2 A B
Displacement is NEVER greater than distance traveled!

24. Scalars vs. Vectors (continued)
Velocity:
has magnitude & direction (example: 15 mi/h North)
Speed:
has a magnitude only (example: 30 km/h) 1 2
Total time for the trip from 1 to 2: 1 hr
25 km
16o
24 km
7 km
Don’t worry about this notation for this test 
Speed = 31 km/h
Velocity = 25 km/h at 16o NE
If an object STARTS & STOPS at the same point, the velocity is ZERO! (since the displacement is zero)

25. x-t graphs
t (sec)
x (m)
t t t3 x2 x1 x3 B C D A
Constant speed (Constant + velocity, or constant velocity in the + direction)
Slow down, speed up, slow down, speed up
2 moments where the object is “at rest” (for a moment)

26. How to get the position (x) at a certain time (t) off an x-t graph
x(m)
t (s) 30 20 10
Example:
What is the position at t = 30 seconds?
24m
Go over to t = 30.
Find the pt on the curve.
Find the x value for this time.

27. How to calculate the displacement between two times on an x-t graph
x(m)
t (s) 30 20 10
Example:
What is the displacement from t = 10 to t = 40?
17 m
Find x1
Find x2
10 m
Use D x = x2 - x1 = + 7 m

28. How to find the distance traveled between two times on an x-t graph.
x(m)
t (s) 30 20 10
Example:
What is the distance traveled from t = 10 to t = 40?
17 m
10 m
Find the distance traveled in the + direction.
Find the distance traveled in the - direction.
Add them together. (27 m)

29. Understand the difference between velocity and speed on an x-t graph.
x(m)
t (s) 30 20 10
Example:
What is the average speed from t = 10 to t = 40 seconds?
17 m
10 m
dist10-40 = 27 m
(previous slide)
Avg. Speed = dist/ Dt
= 27m / 30 sec
= 0.9 m/s

30. Understand the difference between velocity and speed on an x-t graph.
x(m)
t (s) 30 20 10
Example:
What is the average velocity from t = 10 to t = 40 seconds?
Dx10-40 = + 7 m (previous slide)
Avg. Velocity = slope
= Dx/ Dt
= + 7 / 30 sec
= m/s

31. Will avg. velocity EVER be greater than avg. speed?
NO!!!
Will avg. velocity EVER be equal to avg. speed?
YES!!! When the path travelled was one-way, in a straight line.

32. Negative Average Velocity?
x(m)
t (s) 30 20 10
Example:
What is the average velocity from t = 20 to t = 40 seconds?
Avg. vel. = slope = rise/run = -7 m / 20
= -.35 m/s
Since the objects displacement is in the NEGATIVE direction, so is its average velocity.

33. Open to in your Unit 1 packet 1
-10 m 2) 3) 4)
avg velocity = slope = -15m / 6sec = -2.5 m/s
s = |v| = 2.5 m/s
At rest at t = 0 and t = 12 sec

34. Speeding up, const negative vel, slowing down, speeding up, 5) 6)
Speeding up, const negative vel, slowing down, speeding up,
const positive velocity(slow), speeding up, constant positive velocity (fast)
Dx = x2 – x1 = (-10m) – (10m) = -20m (approximately)