What is an Instant of Time?
Car on racetrack example:
1. How long did a car spend at any one
location?
2. Each position is linked to a time, but
how long did that time last?
3. You could say an “instant”, but how long
is that?
CHAPTER – 5
A Mathematical Model of Motion
5.1 Graphing Motion in One Dimension
Position – Time Graphs
Example:
Football running back motion (displacement)
diagram at 1 second intervals.
Plot the Position/Time Graph
4. If an instant lasts for a finite amount of
time, then, because the car would be at
the same position during that time, the car
would be at rest.
But, a moving object (car) cannot be at
rest;
 an instant is not a finite period of time.
5. This means that an instant of time lasts
“0” seconds.
Using a Graph to Find Out Where
and When (Pick various locations)
Graphing the Motion of Two or
More Objects
A = running back
B = linebacker
C = center
D = defensive back
From Graphs to Words and Back
Again
Keep in mind that when t=0, the position of
the object does not necessarily have to be
zero.
Uniform Motion
Definition of uniform motion =
Means that equal displacements occur during
successive equal time intervals.
What does the “slope” of the pos/time
graph give us?
Velocity
rise
Δy
Yf - Yi
slope = run = Δx = xf - xi
Δd df - di
slope = v = Δt = tf - ti
For Objects With Diff. Velocities
Using an Equation to Find out
Where and When
Δd df - di
Average Velocity = v = Δt = tf – ti
d = di + vt
Example Problem (PP-11)
A car starts 200m west of the town square
and moves with a constant velocity of
15m/s towards the east.
a) Where will the car be 10 min later?
b) When will the car reach the town
square?
a) draw sketch
d = di + vt
d = -200m + (15ms)(600s)
d = -200m + 9000m
d = 8800m
b) d = di + vt
0 = -200m + (15m/s)t
200m = (15m/s)t
13.3s = t
Example-2 (PP 12)
A car starts 200m west of the town square
and moves with a constant velocity of
15m/s towards the east. At the same time
a truck was 400m east of the town square
moving west at a constant velocity of
12m/s.
Find the time and place where the car meets
the truck.
Draw a sketch.
d = di + vt
5.2 Graphing Velocity in One
Dimension, Determining
Instantaneous Velocity
Q: When an object is not moving with
uniform motion, the object is said to be…?
A: accelerating
Position-Time Graph
Uniform Motion
Position-Time Graph
Acceleration
Instantaneous Velocity = ?
How fast an object is moving at a
particular instant in time, ie: how fast is it
moving “Right Now”
What is the instantaneous velocity at 2s?
What is the instantaneous velocity at 4s?
Instantaneous Velocity
The Instantaneous Velocity is equal to the
“Slope” of the tangent line of a
position/time graph at any particular time.
Draw previous graph and calculate the
instantaneous velocity at 2s & 4s.
Velocity-Time Graphs
2 planes
Plane-A & Plane-B
vB is a constant 75m/s
vA is constantly increasing (constant “a”)
Draw sketch on board
Q: At the point of intersection, will the
planes crash?
A: ??? Not enough information given, the
graph merely indicates the planes have
the same velocity at that point.
Displacement from a Velocity-Time Graph
Q: What does the area under a V-T graph
represent?
A: Δd, displacement.
5.3 Acceleration
Determining Average Acceleration
Average acceleration, “a” , is equal to the
slope (rise/run, Δv/Δt) of a velocity-time
graph.
Constant and Instantaneous
Acceleration
If there is a constant slope on a VelocityTime Graph then there is also a constant
acceleration, any point “a” is the same.
Acceleration is simply the slope of the line.
Instantaneous Acceleration of a
Velocity-Time Graph, Curved Line
Draw graph on board.
Q: What is the instantaneous acceleration
at 2s? How could it be determined?
A: Draw a tangent line at 2s then calculate
the slope.
Positive and Negative Acceleration
V1
V2
V3
V4
V5
V6
speed increase/decrease/constant
velocity +/acceleration +/-/0
When v+ and a+, speed increases(+)
When v+ and a-, speed decreases(+)
When v- and a-, speed increases(-)
When v- and a+, speed decreases(-)
Calculating Velocity from
Acceleration
v = vo + at
Example Problem
Hines Ward is running for a touchdown at
4m/s. He accelerates for 3s. His velocity
entering the end zone is 7m/s. What is his
acceleration?
v = vo + at
7m/s = 4m/s +a(3s)
3m/s = a(3s)
1m/s2 = a
Displacement Under Constant
Acceleration
d = di + ½(vf + vi)t
d = di + vit + 1/2at2
v2 = vi2 + 2a(df – di)
5.4 Free Fall
Acceleration Due to Gravity
1. Drop a flat sheet of paper
2. Drop a crumpled piece of paper.
3. Drop a tennis ball.
Q: Is there a difference in the acceleration
of each of the objects above?
A: NO All “free falling” objects accelerate
with a magnitude of 9.8m/s2 towards the
center of the Earth.
Acceleration due to gravity “g” = -9.8m/s2
Example: Drop A Rock
After 1s it is falling at ____m/s
After 2s it is falling at ____m/s
After 3s it is falling at ____m/s
After 4s it is falling at ____m/s
 Each second during free fall the rock will
Δ its velocity by -9.8m/s.
Drop a Rock Diagram
DIAGRAM
During each equal successive time interval the
rock will fall a greater distance b/c a = “-g”
Q: Could this diagram also apply to a rock
thrown upward?
A: YES, the diagram would look the same.
Q: Why?
A: Once the rock leaves the hand, the only force
acting on the rock is gravity “g”.
Diagram/Example
Diagram of a ball thrown upward with a
velocity of 49m/s.
Show the velocity changes for 10 seconds,
5 seconds up & 5 seconds down.
DRAW SKETCH
Time
0s
1s
2s
3s
4s
5s
6s
7s
8s
9s
10s
Velocity
49m/s
39.2m/s
29.4m/s
19.6m/s
9.8m/s
0m/s
-9.8m/s
-19.6m/s
-29.4m/s
-39.2m/s
-49m/s
Example Problem
If you throw a rock upward with an initial
velocity of 35m/s:
a) what is its velocity after 1,2,3,4,5 sec?
b) what is its position after 1,2,3,4,5 sec?
c) how long will it take to reach its
maximum height?
d) what is its maximum height?
e) how long will it be in the air?
Example Thrown Upward
Quadratic Equation
y = ax2 + bx + c
When y = 0
0 = ax2 + bx + c
Solving for “x”
- b ± √(b2 – 4ac)
x =
2a
A ball is thrown upward with an initial
velocity of 35m/s, how long will the ball be
in the air?
Equation to be used…?
d = di + vit + 1/2at2
d = ½(a)t2 + vit + di
0 = ½(-9.8m/s2)t2 + (35m/s)t + 0
0 = (-4.9m/s2)t2 + (35m/s)t
Solve for “t”
- b + √(b2 – 4ac)
x =
2a
-35m/s + √﴾(35m/s)2 – 4(-4.9m/s)(0)﴿
t=
2(-4.9m/s2)
-35m/s + √1225m2/s2
t=
-9.8m/s2
-35m/s + 35m/s
0__
t=
-9.8m/s2
= -9.8m/s2 = 0s
OR →
- b - √(b2 – 4ac)
x =
2a
-35m/s - √﴾(35m/s)2 – 4(-4.9m/s)(0)﴿
t=
2(-4.9m/s2)
-35m/s - √1225m2/s2
t=
-9.8m/s2
-35m/s - 35m/s
-70m/s
t=
-9.8m/s2
= -9.8m/s2
t = 7.143s