unit 1 motion graphs lyzinskiphysics x t days 1 - 2
TRANSCRIPT
UNIT 1Motion Graphs
LyzinskiPhysics
x
t
Days 1 - 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.
- MechanicsKINEMATICS
DYNAMICS- Electricity
- Magnetism
- Optics
- Waves
PHYSICSA “description” of motion
A study of what “causes” motion
Day #1
* Distance * Speed * Scalars * d-t graphs
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
Using the number line above, what would be the distance travelled if an object travelled from …..
- A to B
- A to C
- A to C and then back to A
- C to B, passing through A
B C
-3 -2 -1 0 1
Ameters
1m
4m
4m + 4m = 8m
4m + 1m = 5m
5 yd12 yd
A
B
C
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
Definition• Average Speed (s) – the distance travelled
during a time interval divided by the elapsed time.
s = d/t
(or s=d/t) Since t = t2 – t1, if t1 = 0, then t = t2 – 0 = t2 = t
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?
B C
-3 -2 -1 0 1
Amiles
d = 3mi + 4mi = 7mis = 6 mi/h
s = d/t t = d/s = (7mi)/(6mi/h)=1.17h
1 h, 10 minUse appropriate units
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…
B C
-3 -2 -1 0 1
Akm
a) Larry’s average speed during the first portion of the trip.
b) The average speed during the second portion of the trip.
c) 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) = 0.114 km/min = 6.86 km/h
Definition• 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?
d-t graphsConstant speed
Speeding UP
Constant Speed (faster!)
Slowing Down
At rest
t (sec)
d (m)
BC
E
A
D
F
10 15 20 27 36
120
100
50
30
Speed on a d-t graph can be found by taking the _______________.SLOPE
sAB = rise/run = (30-0m) / (10-0s) = 3 m/s
sCD = rise/run = (100-50m) / (20-15s) = 10 m/s
t (sec)
d (m)
BC
E
A
D
F
10 15 20 27 36
120
100
50
30
Open to in your Unit 1 packet
520 – 170yd = 350 yd (approximately)
1) On the d-t graph, find the distance travelled from 4 to 16 seconds. ______
1
d-t graphs CANNOT have sharp points
NOTHING CAN STOP INSTANTANEOUSLY!!
Day #2
* Position* Displacement * Average Velocity* Vectors* x-t graphs
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.
-3 -2 -1 0 1
Akm
6 7 8 9 10
The position of particle “A” is either x = -3 or x = 6, depending on which reference point (or origin) you use.
Definition• Displacement (x) – the change in an
object’s position during a time interval.
x = x2 – x1or
x = 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?
Using the number line above, find the distance travelled and the displacement in moving from
- A to B
- C to A
- A to C and then back to A
- C to B, passing through A
B C
-3 -2 -1 0 1
Ameters
1m, 1m [right]
4m, 4m [left]
8m, 0m
4m, 3m [left]
x = 1 – (1m) = 0m
x = (-2) – (1m) = -3m OR 3m [left]
Definition• Average Velocity ( v ) – the displacement
of an object divided by the elapsed time.
v = x/t
(or v=x/t)
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 speed during each interval are as follows.
AB: 7 m/sBC: 8 m/sCD: 6 m/s
DA: 7.5 m/s
s = d/t t = d/s = 100m/7sec = 14.286 sec100m/8sec = 12.5 sec100m/6sec = 16.667 sec100m/7.5sec = 13.333 sec
s = d/t = (400m)/(56.786s) = 7.04 m/sec
Find Sam’s avg. speed and avg. velocity for the entire trip.
Avg Velocity = 0 since x = 0 for the entire trip.
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, 14.286 secBC: 8 m/s, 12.5 sec
CD: 6 m/s, 16.667 secDA: 7.5 m/s, 13.333 sec
31.831 m
100
104.94
Find Sam’s average speed and average velocity for the 1st half of the race.
s = d/t t = d/s = 200m/(14.286+12.5s) = 7.47 m/s
v = x/t = (104.94m )/(14.286+12.5s) = 3.92 m/sec
Definition• 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
Scalars vs. Vectors
has magnitude & direction (example: 15 cm east)
has a magnitude only (example: 6 ft)
1 2
AB
Displacement is NEVER greater than distance traveled!
Displacement:
Distance:
Scalars vs. Vectors (continued)
has magnitude & direction (example: 15 mi/h North)
has a magnitude only (example: 30 km/h)
If an object STARTS & STOPS at the same point, the velocity is ZERO! (since the displacement is zero)
Velocity:
Speed:
1
2
24 km
7 km
Total time for the trip from 1 to 2: 1 hr
Speed = 31 km/h
Velocity = 25 km/h at 16o NE
25 km
16o Don’t worry about this notation for this test
x-t graphs
t (sec)
x (m)
t1 t2 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)
How to get the position (x) at a certain time (t) off an x-t graph
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Example:
What is the position at t = 30 seconds?
Go over to t = 30.
Find the pt on the curve.
Find the x value for this time.
24m
How to calculate the displacement between two times on an x-t graph
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Example:
What is the displacement from t = 10 to t = 40?
Find x1
Find x2
Use x = x2 - x1 = + 7 m 10 m
17 m
How to find the distance traveled between two times on an x-t graph.
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Example:
What is the distance traveled from t = 10 to t = 40?
Find the distance traveled in the + direction.
Find the distance traveled in the - direction.
Add them together. (27 m)
17 m10 m
Understand the difference between velocity and speed on an x-t graph.
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Example:
What is the average speed from t = 10 to t = 40 seconds?
dist10-40 = 27 m
(previous slide)
Avg. Speed = dist/ t
= 27m / 30 sec
= 0.9 m/s
17 m
10 m
Understand the difference between velocity and speed on an x-t graph.
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Example:
What is the average velocity from t = 10 to t = 40 seconds?
Avg. Velocity = slope
= x/ t
= + 7 / 30 sec
= + 0.23 m/s
x10-40 = + 7 m (previous slide)
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.
Negative Average Velocity?
x(m)
10 20 30 40 50
t (s)
30
20
10
0
Avg. vel. = slope = rise/run = -7 m / 20
= -.35 m/s
Example:
What is the average velocity from t = 20 to t = 40 seconds?
Since the objects displacement is in the NEGATIVE direction, so is its average velocity.
1) On the x-t graph, find the position at 9 seconds. ______
2) On the x-t graph, find the average speed AND the average velocity from 6 to 12 seconds.
3) On the x-t graph, name all the times (or time intervals) during which the object is at rest.
-10 m
avg velocity = slope = -15m / 6sec = -2.5 m/s s = |v| = 2.5 m/s
At rest at t = 0 and t = 12 sec
Open to in your Unit 1 packet1
2)
3)
4)
1) On the x-t graph, name each different motion interval (hint: there are 7 answers).
2) On the x-t graph, find the displacement during the 1st 21 seconds.
x = x2 – x1 = (-10m) – (10m) = -20m (approximately)
5)
6)
Speeding up, const negative vel, slowing down, speeding up, const positive velocity(slow), speeding up, constant positive velocity (fast)