motion in one dimension (velocity/speed vs. time) chapter 5.2
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Motion in One Dimension(Velocity/Speed vs. Time)
Chapter 5.2
What is instantaneous speed?
distance time speed time speed
(m) (s) (m/s) (s) (m/s)0 0 0 0 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
100M Sprint Final
World Championships; Athens, Greece; 1997
Split times and instantaneous speeds in 10 m intervals
Maurice Greene Donovan Bailey
What effect does an increase in speed have on displacement?
d1
d2
d3
d4
d5
d6
Instead of position vs. time, consider velocity or speed vs. time.
High acceleration
Relatively constant speed = no 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:
Therefore, slope must equal acceleration.
Time
t
v
x
ySlope
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
Zero Acceleration
C
Time
Negative Acceleration
B
Acceleration determined from the slope of the curve.
rise run
vf – vi
tf – ti
8.4m/s-0m/s 1.7s-0.00s
m = 4.9 m/s2
Since m = a:
a = 4.9 m/s2
m =
Slope =
m =
What is the acceleration from t = 0 to t = 1.7 seconds?
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
Time
A2A1
A1 = d1 = ½ v1*t1
A2 = d2 = v2*t2
dtotal = d1 + d2
Measuring displacement from a velocity vs. time graph.
A = ½ b x hA = ½ (2.36s)(11.7m/s)A = 13.8 m
A = b x hA = (7.37s)(11.7m/s)A = 86.2 m
Algebraically deriving the kinematics formulas in your reference table
Determining velocity from acceleration
If acceleration is considered constant:
• Since ti is normally set to 0, this term can be eliminated.
• Rearranging terms to solve for vf results in:
Time
What is the average velocity? If the total displacement and time are known, the average
velocity can be found using the formula:
However, if all you have are the initial and final velocities, the
average value can be found by:
Time
This latter formula is not in your reference table!
𝑣 𝑖
𝑣 𝑓
𝑣𝑎𝑣𝑔
How to determine position, velocity or acceleration without time.
(1)
(2)
Solve (2) for t: and substitute back into (1)
By rearranging to solve for :
(3)
2
)( ifavg
vvv
d
Remember, since
Note that
How to determine displacement, time or initial velocity without the final velocity.
(1)
(2)
Substitute (2) into (1) for vf
(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).
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
vfn/a
vi 0 m/s
t 3.00 s
Example 1: Calculating Distance
From your reference table:
Since vi = 0 we will substitute g for a and y for d to get:
Example 2: Calculating Final Velocity What will the final velocity of the stone be?
Data
y -44.1 m
a = g -9.81 m/s2
vf?
vi 0 m/s
t 3.00 s
Example 2: Calculating Final Velocity Using your reference table:
Again, since vi = 0 and substituting g for a, we get:
Or, we can also solve the problem with:
Example 3: Determining the Maximum Height How high will the coin go?
Data
y ?
a = g -9.81 m/s2
vf0 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:
Since Δd = Δy we can algebraically rearrange the terms to solve for Δy.
2 2
2f iv v
yg
2 2
2
(0 / ) (5 / )1.27
2( 9.81 / )
m s m sy m
m s
Example 4: Determining the Total Time in the Air How long will the coin be in the air?
Data
y 1.27 m
a = g -9.81 m/s2
vf0 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:, where
Solving for t gives us:
Since the coin travels both up and down, this value must be doubled to get a total time of 1.02s
f iv vt
g
2
0 / 5 /0.510
9.81 /
m s m st s
m s
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 = vi
2 + 2a*(df – di) vf = vi +at a = Δv/Δt = (vf – vi)/(tf – ti)
Determining instantaneous velocity
1997 World Championships - Athens, GreeceMaurice Green
0
10
20
30
40
50
60
70
80
90
100
0 2 4 6 8 10
Time (s)
Dis
tan
ce (
m)
y = 1.13x + 4.08x - 0.05
R2 = 1.00
2
y = 11.65x - 13.07
R = 1.002
How do you determine the instantaneous velocity?
Instantaneous velocity = slope of line tangent to curve.
What is the runners velocity at t = 1.5s?
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.
rise run
vf – vi
tf – ti
13m/s-7m/s 3.75s-0.75s
m = 2.0 m/s2
Since m = a:
a = 2.0 m/s2
m =
Slope =
m =
What is the acceleration at t = 2 seconds?
Displacement when acceleration is constant.
Displacement Under Constant Acceleration
0
10
20
30
40
50
60
70
80
90
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Vel
oci
ty (
m/s
)
Displacement = area under the curve.
Δd = vit + ½ (vf – vi)*t
Simplifying:
Δd = ½ (vf + vi)*t
If the initial position, di, is not 0, then:
df = di + ½ (vf + vi)*t
By substituting vf = vi + at
df = di + ½ (vi + at + vi)*t
Simplifying:
df = di + vit + ½ at2
d = vit
d = ½ (vf-vi)t
vf
vi
t