Physics Intro & Kinematics
Some Physics Quantities
Vector - quantity with both magnitude (size) and direction
Scalar - quantity with magnitude only
Vectors:
Scalars:
Mass vs. Weight
On the moon, your mass would be the same, but the magnitude of your weight would be less.
Mass
Weight
Vectors
Vectors are represented with arrows
42°
5 m/s
Units
Quantity . . . Unit (symbol)
Units are not the same as quantities!
SI Prefixes
Little Guys
Big Guys
Kinematics definitions
Distance vs. Displacement
start
stop
Speed, Velocity, & Acceleration
Speed vs. Velocity
Speed vs. Velocity
Acceleration
Acceleration – how fast you speed up, slow down, or change direction; it’s the rate at which velocity changes. Two examples:
t (s) | v (mph) |
0 | 55 |
1 | 57 |
2 | 59 |
3 | 61 |
t (s) | v (m/s) |
0 | 34 |
1 | 31 |
2 | 28 |
3 | 25 |
a = +2 mph / s
a = -3
m / s
s
= -3 m / s 2
Velocity & Acceleration Sign Chart
| V E L O C I T Y | ||
ACCELERATION | | + | - |
+ | Moving forward;�Speeding up | Moving backward;�Slowing down | |
�- | Moving forward;�Slowing down | Moving backward;�Speeding up |
Acceleration due to Gravity
a = -g = -9.8 m/s2
9.8 m/s2
Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance).
Interpretation: Velocity decreases by 9.8 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down.
This acceleration vector is the same on the way up, at the top, and on the way down!
Kinematics Formula Summary
(derivations to follow)
For 1-D motion with constant acceleration:
Kinematics Derivations
a = Δv / Δt (by definition)
a = (vf – v0) / t
⇒ vf = v0 + a t
vavg = (v0 + vf ) / 2 will be proven when we do graphing.
Δx = v t = ½ (v0 + vf) t = ½ (v0 + v0 + a t) t� ⇒ Δx = v0 t + a t 2
(cont.)
Kinematics Derivations (cont.)
vf = v0 + a t ⇒ t = (vf – v0) / a�Δx = v0 t + a t 2 ⇒� Δx = v0 [(vf – v0) / a] + a [(vf – v0) / a] 2 � ⇒ vf2 – v02 = 2 a Δx
Note that the top equation is solved for t and that expression for t is substituted twice (in red) into the �Δx equation. You should work out the algebra to prove the final result on the last line.
Sample Problems
Multi-step Problems
19.8 m/s
188.83 m
Answer:
Answer:
Graphing !
x
t
A
B
C
A … Starts at home (origin) and goes forward slowly
B … Not moving (position remains constant as time progresses)
C … Turns around and goes in the other direction � quickly, passing up home
1 – D Motion
Graphing w/ Acceleration
x
A … Start from rest south of home; increase speed gradually
B … Pass home; gradually slow to a stop (still moving north)
C … Turn around; gradually speed back up again heading south
D … Continue heading south; gradually slow to a stop near the � starting point
t
A
B
C
D
Tangent Lines
t
SLOPE | VELOCITY |
Positive | Positive |
Negative | Negative |
Zero | Zero |
SLOPE | SPEED |
Steep | Fast |
Gentle | Slow |
Flat | Zero |
x
On a position vs. time graph:
Increasing & Decreasing
t
x
Increasing
Decreasing
On a position vs. time graph:
Increasing means moving forward (positive direction).
Decreasing means moving backwards (negative direction).
Concavity
t
x
On a position vs. time graph:
Concave up means positive acceleration.
Concave down means negative acceleration.
Special Points
t
x
P
Q
R
Inflection Pt. | P, R | Change of concavity |
Peak or Valley | Q | Turning point |
Time Axis Intercept | P, S | Times when you are at “home” |
S
Curve Summary
t
x
A
B
C
D
All 3 Graphs
t
x
v
t
a
t
Graphing Animation Link
This website will allow you to set the initial velocity and acceleration of a car. As the car moves, all three graphs are generated.
Graphing Tips
t
x
v
t
Graphing Tips
The same rules apply in making an acceleration graph from a velocity graph. Just graph the slopes! Note: a positive constant slope in blue means a positive constant green segment. The steeper the blue slope, the farther the green segment is from the time axis.
a
t
v
t
Real life
Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the green segments would be connected. In our class, however, we’ll mainly deal with constant acceleration.
a
t
v
t
Area under a velocity graph
v
t
“forward area”
“backward area”
Area above the time axis = forward (positive) displacement.
Area below the time axis = backward (negative) displacement.
Net area (above - below) = net displacement.
Total area (above + below) = total distance traveled.
Area
The areas above and below are about equal, so even though a significant distance may have been covered, the displacement is about zero, meaning the stopping point was near the starting point. The position graph shows this too.
v
t
“forward area”
“backward area”
t
x
Area units
v (m/s)
t (s)
12 m/s
0.5 s
12
Graphs of a ball thrown straight up
x
v
a
The ball is thrown from the ground, and it lands on a ledge.
The position graph is parabolic.
The ball peaks at the parabola’s vertex.
The v graph has a slope of -9.8 m/s2.
Map out the slopes!
There is more “positive area” than negative on the v graph.
t
t
t
Graph Practice
Try making all three graphs for the following scenario:
1. Schmedrick starts out north of home. At time zero he’s driving a cement mixer south very fast at a constant speed.
2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose.
3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail.
4. Fleeing the scene of the crime, Schmedrick takes off again in the same direction, speeding up quickly.
5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.
Kinematics Practice
A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop? Be mindful of your units!
2.24 ms
Answer
Uniform Acceleration
When object starts from rest and undergoes constant acceleration:
t : 0 1 2 3 4
Δx = 1
Δx = 3
Δx = 5
( arbitrary units )
x : 0 1 4 9 16
Δx = 7
Spreadsheet Problem
Relationships
Let’s use the kinematics equations to answer these:
1. A mango is dropped from a height h.
a. If dropped from a height of 2 h, would the impact speed double?
Relationships (cont.)
3 v
9 v2 / 2 g
6 v / g
Answers