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Physics Unit One. Motion. Motion. Any change in the position of an object. Can be completely described with speed and direction. The distance an object travels per unit of time. s = d / t Units - any distance over any time. Ex. m/s, km/hr, cm/s, etc…. Speed PLUS direction. - PowerPoint PPT Presentation
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Physics Unit One
Motion
Motion
• Any change in the position of an object.
• Can be completely described with speed and direction.
Speed vs. Velocity
• The distance an object travels per unit of time.
• s = d / t• Units - any distance
over any time.
Ex. m/s, km/hr, cm/s, etc…
• Speed PLUS direction.• A vector quantity
(magnitude + direction)• v = d / t, direction• Units – any distance
over any time, plus, direction.
Ex. 6.5 cm/s, due north
5-Step Method for Solving Problems
A cat can run 5.0 meters in 3.25 seconds. What is the cat’s average speed?
d = 5.0 m
t = 3.25 s
s = d / t
s = 5.0m / 3.25s
s = 1.54 m/s
Velocity Sample Problem
• A tennis ball flies off the end of a racket and travels 10.0 meters in 0.95s. What is the velocity of the tennis ball?
d = 10.0 mt = 0.95s
v = d / tv = 10.0m / 0.95sv = 10.53 m/s, across the net
d vs. t graphs
• Graph the motion
• use the slope to interpret the relationship between the variables (d & t)
Slope =riserun = =
ΔYΔX
dt
= Speed
faster the speed
slower the speed
*steeper the slope
*more gentle the slope
d
t
d d d
t tt
When interpreting the motion answer:
•As time goes by, distance (d or )
•The line shows relationship (constant or changing)
•Moving forward (d)
•Constant speed (straight)
•Moving backward (d)
•Moving forward (d)
•Constant speed (straight)
•Constant speed (straight)
•No motion (d stays same)
•0 m/s
•Changing speed (curved)•positive acceleration
d dd
t t t•Moving forward (d)
•Moving backward (d)
•Moving backward (d)
•changing speed (curved)
•changing speed (curved)
•changing speed (curved)
•(-) acceleration •(+) acceleration•(-) acceleration
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
Dis
tanc
e (m
)
A
B C
D
E
H
G
F
AB
•Move forward•Constant speed•ΔYΔX
=10m10s =1m/s
BC•At rest
•Constant speed•0m/s
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
Dis
tanc
e (m
)
A
B C
D
E
H
G
F
CD
•Moving backward•Constant speed•ΔYΔX
= 5m 5s
=1m/s
DE•Moving forward
•Changing speed•(+) acceleration
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
Dis
tanc
e (m
)
A
B C
D
E
H
G
F
EF•Moving backward•Constant speed•ΔYΔX
=25m 5s
=5m/s
FG•Moving forward•Constant speed•ΔYΔX
==10m5s
2m/s
GH•Moving forward•Constant speed•ΔYΔX
= =12.5m
12.5s1m/s
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
Dis
tanc
e (m
)
A
B C
D
E
H
G
F
Which line had the fastest speed?•EF•Steepest slope
Distance increases?
Distance decreases?•Moving forward
•Moving backward
Curved line?
Straight line?•Constant speed
•Changing speed
d vs. t graph shows?•Speed and direction
Unit 1 Physics
Acceleration
Acceleration
• Any change in velocity (speed or direction) of an object
a = vf - vi
ta = acceleration
vf = final velocityvi = initial velocity
t = time
Example:A runner increases her speed from 3m/s to 10m/s in 2 seconds. Calculate her rate of acceleration.
vf = 10m/s
vi = 3m/s
t = 2s
a = vf - vi
ta = 10m/s - 3m/s 2s
a = 7m/s 2sa = 3.5m/s2
She gets 3.5m/s faster every second.
0s = 3m/s 1s = 6.5m/s 2s = 10m/s
Example: A car goes from 88km/hr to stopped in 4s.
vf = 0km/hr
vi = 88km/hr
t = 4sa = vf - vi
t
a = 0km/hr - 88km/hr 4sa = -88km/hr 4s
a = -22km/hr s
The car gets 22km/hr slower every second.
0s = 88km/hr
1s = 66km/hr
2s = 44km/hr
3s = 22km/hr
4s = 0km/hr
v
t
v v v
t tt
When interpreting the motion answer:
•As time goes by, velocity (v or )
•The line shows relationship (constant or changing)
•Constantpositiveacceleration
•Constantnegativeacceleration
•Constantacceleration0m/s2
•Constant speed
•Changingpositiveacceleration
v vv
t t t•Changingpositive acceleration
•Changingnegativeacceleration
•Changingnegativeacceleration
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
velo
city
(m
/s)
A
B C
D
E
H
G
F
AB•Constant (+) acceleration
10m/s10s
= 1m/s2
BC•Constant speed10m/s
•Constant acceleration 0m/s2
10m/s - 0m/s = 10s
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
velo
city
(m
/s)
A
B C
D
E
H
G
F
CD•Constant (-) acceleration
-5m/s5s
= -1m/s2
DE•Changing (+) acceleration
5m/s - 10m/s = 5s
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
velo
city
(m
/s)
A
B C
D
E
H
G
F
EF•Constant (-) acceleration
-25m/s5s
= -5m/s2
FG•Constant (+) acceleration
0m/s - 25m/s = 5s
10m/s - 0m/s = 5s
10m/s5s
= 2m/s2
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
velo
city
(m
/s)
A
B C
D
E
H
G
F
GH•Constant (+) acceleration
12.5m/s12.5s
= 1m/s2
22.5m/s - 10m/s = 12.5s
MOTION
-
Time (s)
0 5 10 15 20 25 30 35 40 45 50 55
5
10
15
20
25
30
velo
city
(m
/s)
A
B C
D
E
H
G
F
The car accelerated,
stayed at the same speed,
slowed down for light,
light turns & quickly accelerates,
slams on the brakes to stop for red light,
quickly accelerates,
continues at slower acceleration
