Motion in One Dimension (Velocity vs. Time) Chapter 5.2

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Motion in One Dimension(Velocity vs.
Time)Chapter 5.2
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What is instantaneous velocity?
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What effect does an increase in velocity have on
displacement?
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Instead of position vs. time, consider velocity
vs. time.
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How can be determined from a v vs. t graph?

• Measure the under the curve.
• d vt
• Where
• t is the x component
• v is the y component

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Measuring displacement from a velocity vs. time
graph.
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What information does the slope of the velocity
vs. time curve provide?

1. sloped curve velocity (Speeding
   up).
2. sloped curve velocity (Slowing
   down).
3. sloped curve velocity.

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What is the significance of the slope of the
velocity 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 .

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Acceleration determined from the slope of the
curve.
What is the acceleration from t 0 to t 1.7
seconds?
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Determining velocity from acceleration

• If acceleration is considered constant
• a ?__/?__ (__ __)/(__ __)
• Since ti is normally set to 0, this term can be
  eliminated.
• Rearranging terms to solve for vf results in
• __ __ a__

Velocity
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Position, velocity and acceleration when t is
unknown.

• __ __ ½ (vf vi)t (1)
• vf vi at (2)
• Solve (2) for t t (__ __)/__ and substitute
  back into (1)
• df di ½ (vf vi)(__ __)/__
• By rearranging
• __ __ 2__(_____) (3)

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Alternatively, If time and acceleration are
known, but the final velocity is not

• df di ½ (vf vi)t (1)
• vf vi at (2)
• Substitute (2) into (1) for vf
• df di ½ (__ __ __ __)t
• df di __ __ __ __ __ (4)

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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 ½ vf ?t
vf vi a?t vf a?t
?d vi ?t ½ a(?t)2 ?d ½ a(?t)2
vf2 vi2 2a?d vf2 2a?d
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Acceleration due to

• All falling bodies accelerate at the rate when
  the effects of due to can be ignored.
• Acceleration due to is caused by the
  influences of Earths on objects.
• The acceleration due to is given the special
  symbol .
• The acceleration due to is a to the
  surface of the earth.
• __/__

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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 Data
y ?
a g -9.81 m/s2
vf n/a
vi 0 m/s
t 3.00 s
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Example 1 Calculating Distance

• From your reference table
• d ____ ____
• Since vi __ we will substitute __ for __ and __
  for __ to get
• __ ____

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Example 2 Calculating Final Velocity

• What will the final velocity of the stone be?

Data Data
y -44.1 m
a g -9.81 m/s2
vf ?
vi 0 m/s
t 3.00 s
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Example 2 Calculating Final Velocity

• Using your reference table
• vf __ __ __
• Again, since vi __ and substituting __ for __,
  we get
• vf __ __
• vf
• vf
• Or, we can also solve the problem with
• vf2 ___ _____, where vi __
• vf

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Example 3 Determining the Maximum Height

• How high will the coin go?

Data Data
y ?
a g -9.81 m/s2
vf 0 m/s
vi 5.00 m/s
t ?
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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 ?__ ?__ we can algebraically rearrange
  the terms to solve for ?__.

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Example 4 Determining the Total Time in the Air

• How long will the coin be in the air?

Data Data
y 1.28 m
a g -9.81 m/s2
vf 0 m/s
vi 5.00 m/s
t ?
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Example 4 Determining the Total Time in the Air

• Since we know the and velocity as well as
  the rate of we can use
• __ __ __ __, where __ __
• Solving for t gives us
• Since the coin travels both and , this value
  must be to get a total time of s

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Key Ideas

• Slope of a velocity vs. time graphs provides an
  objects .
• The area under the curve of a velocity vs. time
  graph provides the objects .
• Acceleration due to gravity is the for all
  objects when the effects of due to wind,
  water, etc can be ignored.

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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 ?v/?t (vf vi)/(tf ti)

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Displacement when acceleration is constant.
Displacement area under the curve. ?d vit ½
(vf vi)t Simplifying ?d 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