Motion in One Dimension
(Velocity/Speed vs. Time)
Chapter 5.2
What is instantaneous speed?
100M Sprint Final
Split times and instantaneous speeds in 10 m intervals
World Championships; Athens, Greece; 1997
Maurice Greene
distance
Donovan Bailey
(m)
0
time
(s)
0
speed
(m/s)
0
time
(s)
0
speed
(m/s)
0
10
1.71
8.71
1.78
8.9
20
2.75
10.47
2.81
10.55
30
3.67
11.14
3.72
11.28
40
4.55
11.5
4.59
11.63
50
5.42
11.67
5.44
11.76
60
6.27
11.8
6.29
11.8
70
7.12
11.68
7.14
11.7
80
7.98
11.57
8
11.55
90
8.85
11.51
8.87
11.38
100
9.73
11.3
9.77
11
What effect does an increase in speed
have on displacement?
d6
d5
d4
d3
d2
d1
Instead of position vs. time, consider
velocity or speed vs. time.
Relatively constant
speed = no acceleration
High acceleration
What is the significance of the slope
of the velocity/speed vs. time curve?

Since velocity is on the y-axis and time is on the x-axis, it
follows that the slope of the line would be:
y v
Slope 

x t

Therefore, slope must equal acceleration.
Time
What information does the slope of
the velocity vs. time curve provide?
A. Positively sloped curve = increasing velocity (Speeding
up).
B. Negatively sloped curve = decreasing velocity (Slowing
down).
C. Horizontally sloped curve = constant velocity.
Time
Positive Acceleration
A
Time
Time
Negative Acceleration
Zero Acceleration
B
C
Acceleration determined from
the slope of the curve.
What is the acceleration from t = 0 to t = 1.7 seconds?
rise
run
vf – vi
tf – ti
Slope =
m=
m=
8.4m/s-0m/s
1.7s-0.00s
m = 4.9 m/s2
Since m = a:
a = 4.9 m/s2
How can displacement be determined from
a v vs. t graph?

Measure the area under the curve.
• d = v*t
Where
• t is the x component
• v is the y component
A1 = d1 = ½ v1*t1
A1
A2
A2 = d2 = v2*t2
dtotal = d1 + d2
Time
Measuring displacement from a velocity
vs. time graph.
A=bxh
A = (7.37s)(11.7m/s)
A = 86.2 m
A=½bxh
A = ½ (2.36s)(11.7m/s)
A = 13.8 m
Algebraically deriving the kinematics
formulas in your reference table
Determining velocity from
acceleration

If acceleration is considered constant:
𝑎 = 𝑣/𝑡 = (𝑣𝑓 – 𝑣𝑖 )/(𝑡𝑓 – 𝑡𝑖)
•
•
Since ti is normally set to 0, this term can be eliminated.
Rearranging terms to solve for vf results in:
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
Time
What is the average velocity?

If the total displacement and time are known, the average
velocity can be found using the formula:
𝑣 =

𝑑
𝑡
However, if all you have are the initial and final velocities, the
average value can be found by:
𝑣𝑓
𝑣 =
𝑣𝑖 +𝑣𝑓
𝑣𝑎𝑣𝑔
2
This latter formula is not in your reference
table!
𝑣𝑖
Time
How to determine position, velocity or
acceleration without time.
𝑑𝑓 = 𝑑𝑖 + ½ (𝑣𝑓 + 𝑣𝑖) ∗ 𝑡
(1)
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
(2)
(𝑣𝑓 – 𝑣𝑣
)
𝑖
Solve Remember,
(2) for t: 𝑡 =since
𝑎
𝑑
=
and
𝑡
vavg 
2
= 𝑣𝑡into (1)
substitute𝑑back
½ (𝑣𝑓 + 𝑣𝑖 )(𝑣𝑓 – 𝑣𝑖 )
𝑑𝑓 = 𝑑𝑖 +
𝑎
2
2
Note thatto𝑣solve
By rearranging
𝑓 + 𝑣𝑖for𝑣𝑣
𝑓 𝑓–2 :𝑣𝑖 = 𝑣𝑓 − 𝑣𝑖
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎 ∗ (𝑑𝑓 – 𝑑𝑖)
(v f  vi )
(3)
d
How to determine displacement, time or
initial velocity without the final velocity.
𝑑 = ½ (𝑣𝑓 + 𝑣𝑖) ∗ 𝑡
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
(1)
(2)
Substitute (2) into (1) for vf
𝑑 = ½ (𝑣𝑖 + 𝑎𝑡 + 𝑣𝑖) ∗ 𝑡
𝑑 = 𝑣𝑖𝑡 + ½ 𝑎𝑡2
(4)
Formulas for Motion of Objects
Equations to use when
an accelerating object
has an initial velocity.
Form to use when
accelerating object starts
from rest (vi = 0).
𝑑 = ½(𝑣𝑖 + 𝑣𝑓)𝑡
𝑑 = ½𝑣𝑓𝑡
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
𝑣𝑓 = 𝑎𝑡
𝑑 = 𝑣𝑖𝑡 + ½ 𝑎𝑡2
𝑑 = ½ 𝑎𝑡2
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑑
𝑣𝑓2 = 2𝑎𝑑
Acceleration due to Gravity





All falling bodies accelerate at the same rate when the
effects of friction due to water, air, etc. can be ignored.
Acceleration due to gravity is caused by the influences of
Earth’s gravity on objects.
The acceleration due to gravity is given the special
symbol g.
The acceleration of gravity is a constant close to the
surface of the earth.
g = 9.81 m/s2
Example 1: Calculating Distance

A stone is dropped from the top of a tall building.
After 3.00 seconds of free-fall, what is the
displacement, y of the stone?
Data
y
a=g
?
-9.81 m/s2
vf
n/a
vi
0 m/s
t
3.00 s
Example 1: Calculating Distance

From your reference table:
𝑑 = 𝑣𝑖𝑡 + ½ 𝑎𝑡2

Since vi = 0 we will substitute g for a and
y for d to get:
𝑦 = ½ 𝑔𝑡2
𝑦 = ½ (−9.81 𝑚/𝑠2)(3.00 𝑠)2
𝑦 = −44.1 𝑚
Example 2: Calculating Final
Velocity

What will the final velocity of the stone be?
Data
y
a=g
-44.1 m
-9.81 m/s2
vf
?
vi
0 m/s
t
3.00 s
Example 2: Calculating Final
Velocity



Using your reference table:
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
Again, since vi = 0 and substituting g for a, we get:
𝑣𝑓 = 𝑔𝑡
𝑣𝑓 = (−9.81 𝑚/𝑠2)(3.00 𝑠)
𝑣𝑓 = −29.4 𝑚/𝑠
Or, we can also solve the problem with:
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑑, 𝑤ℎ𝑒𝑟𝑒 𝑣𝑖 = 0
𝑣𝑓 = [(2(−9.81 𝑚/𝑠2)(−44.1 𝑚)]1/2
𝑣𝑓 = −29.4 𝑚/𝑠
Example 3: Determining the
Maximum Height

How high will the coin go?
Data
y
a=g
?
-9.81 m/s2
vf
0 m/s
vi
5.00 m/s
t
?
Example 3: Determining the
Maximum Height


Since we know the initial and final velocity as well as
the rate of acceleration we can use:
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑑
Since Δd = Δy we can algebraically rearrange the
terms to solve for Δy.
y
v 2f  vi2
2g
(0m / s ) 2  (5m / s) 2
y
 1.27 m
2
2(9.81m / s )
Example 4: Determining the
Total Time in the Air

How long will the coin be in the air?
Data
y
a=g
1.27 m
-9.81 m/s2
vf
0 m/s
vi
5.00 m/s
t
?
Example 4: Determining the
Total Time in the Air

Since we know the initial and final velocity as well as
the rate of acceleration we can use:
𝑣𝑓 = 𝑣𝑖 + 𝑎Δ𝑡, where 𝑎 = 𝑔
Solving for t gives us:
vf  vi
t
g
0m / s  5m / s
t
 0.510s
2
9.81m / s

Since the coin travels both up and down, this value
must be doubled to get a total time of 1.02s
Key Ideas




Instantaneous velocity is equal to the slope of a line tangent
to a position vs. time graph.
Slope of a velocity vs. time graphs provides an objects
acceleration.
The area under the curve of a velocity vs. time graph
provides the objects displacement.
Acceleration due to gravity is the same for all objects when
the effects of friction due to wind, water, etc can be ignored.
Important equations to know for
uniform acceleration.
df = di + ½ (vi + vf )*t
 df = di + vit + ½ at2
 vf2 = vi2 + 2a*(df – di)
 vf = vi +at
 a = Δv/Δt = (vf – vi)/(tf – ti)

Determining instantaneous
velocity
1997 World Championships - Athens, Greece
Maurice Green
100
90
80
y = 11.65x - 13.07
2
R = 1.00
Distance (m)
70
60
50
40
30
20
2
y = 1.13x + 4.08x - 0.05
2
R = 1.00
10
0
0
2
4
6
Time (s)
8
10
How do you determine the
instantaneous velocity?
What is the runners
velocity at t = 1.5s?
Instantaneous velocity =
slope of line tangent to
curve.
Determining the instantaneous velocity from
the slope of the curve.
m = rise/run
m = 25m – 5 m
3.75s – 1.0s
m = 7.3 m/s
v = 7.3 m/s @ 1.5s
Acceleration determined from
the slope of the curve.
What is the acceleration at t = 2 seconds?
rise
run
vf – vi
tf – ti
Slope =
m=
m=
13m/s-7m/s
3.75s-0.75s
m = 2.0 m/s2
Since m = a:
a = 2.0 m/s2
Displacement when acceleration is
constant.
Displacement = area under
the curve.
80
Simplifying:
70
Δd = ½ (vf + vi)*t
df = di + ½ (vf + vi)*t
60
Velocity (m/s)
If the initial position, di, is not
0, then:
d = ½ (vf-vi)t
50
40
By substituting vf = vi + at
30
df = di + ½ (vi + at + vi)*t
20
Simplifying:
10
df = di + vit + ½ at2
vf
90
Δd = vit + ½ (vf – vi)*t
Displacem ent Under Constant Acceleration
vi
d = vit
0
0
0.5
1
1.5
2
2.5
Tim e (s)
3
3.5
4
4.5
t
5