Chapter 2 Examples

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Chapter 2 Examples
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Problem 2.46

• An egg is thrown vertically upward from a point
  near the cornice of a tall building. It just
  misses the cornice on the way down and passes a
  point 50 m below its starting point 5.0 s after
  it leaves the throwers hand. Ignore air
  resistance.
• What is the initial speed of the egg?
• How high does it rise above its starting point?
• What is the magnitude of its velocity at its
  highest point?
• What are the magnitude and direction of its
  acceleration at the highest point?

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Step 1 Draw it
h, height above building
50 m in 5 s
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Make some definitions

• Let t5 s
• Let vo initial velocity, in positive direction
• Let g9.8 m/s2 , acceleration due to gravity, in
  negative direction.
• total distance -50 m

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Apply Definitions and Solve

• Let t5 s
• Let vo initial velocity, in positive direction
• Let g9.8 m/s2 , acceleration due to gravity, in
  negative direction. This will be a
• total distance -50 m y(t)-yo

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Part B) Find height above bldg

• v014.5 m/s
• At top of the arc, vf0
• At t0, let y00

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Part B) Find height above bldg

• v014.5 m/s
• At top of the arc, vf0
• At t0, let y00
• Let yfh

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Parts C) and D)

• At the top of the arc, v0 m/s
• Always in this problem a-g-9.8 m/s2

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Problem 2.61

• A car 3.5 m in length and traveling at a constant
  speed of 20 m/s is approaching an intersection
  which is 20 m wide. The light turns yellow when
  the front of the car is 50 m from the beginning
  of the intersection. If the driver steps on the
  brake, the car will slow at -3.8 m/s2. If the
  driver steps on the gas pedal, the car will
  accelerate at 2.3 m/s2. The light will be yellow
  for 3 seconds. Ignore reaction time. To avoid
  being in the intersection when the light changes
  to red, accelerate or brake?

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Step 1 Draw It!
3.5 m
In order for the car to run the green light, It
must traverse 3.55020 m73.5 m so that no part
of the car is in the intersection Braking is
easier, it must traverse less than 50 m
50 m
20 m
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Our Options

• We know that
• t3 s
• a is either 2.3 m/s2 or -3.8 m/s2
• vo20 m/s
• Total distance is either 73.5 m or 50 m

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Our Options

• We know that
• t3 s
• a is either 2.3 m/s2 or -3.8 m/s2
• vo20 m/s
• Total distance is either gt73.5 m or lt50 m

Braking
Run It !
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Problem 2.77

• A physics student with too much free time drops a
  water melon from the roof of a building. He
  hears the sound of the watermelon going splat
  2.5 s later. How high is the building? The
  speed of sound is 340 m/s and ignore air
  resistance.

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Step 1 Draw It!
h
Splat!
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Some Hard Thinkin

• The melon experiences an acceleration due to
  gravity. The student merely dropped it, so its
  initial velocity was 0.
• The sound wave is unaffected by gravity so it
  moves with constant velocity from the ground
  toward the student.
• These are two separate events with a total time
  of 2.5 s

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Some Hard Thinkin part 2

• The equation for the distance that the melon
  traverses is y-1/2g(t1)2 where y height of
  bldg and t1 is the time for the fall.
• The equation of distance for the sound wave is
  yvst2 where vs speed of sound 340 m/s
• The total time for all this to transpire is 2.5 s
  or 2.5 s t1t2

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Solving
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Problem 2.89

• A painter is standing on scaffolding that is
  raised at a constant speed. As he travels
  upward, he accidentally nudges a paint can off
  the scaffolding and it falls 15 m to the ground.
  You are watching and measure with your stopwatch
  that it takes 3.25 s for the can to reach the
  ground.
• What is the speed of the can just before it hits
  the ground?
• Another painter is standing on a ledge with his
  hands 4 m above the can when it falls. He has
  lightning-fast reflexes and can catch if at all
  possible. Does he get a chance?

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Part a) Make some definitions

• Let t3.25 s
• Let vo initial velocity, in positive direction
• Let g9.8 m/s2 , acceleration due to gravity, in
  negative direction.
• total distance -15 m

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Make some definitions

• We must solve this equation for vo

• And then we must use this equation to solve for
  the velocity

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Solving
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Part B) Make some definitions

• The word falls is slightly misleading, the can
  first rises in the air and then falls to the
  ground.
• What it is asking for is how high the can flies
  above the release point it must be greater than
  4 m
• So we are solving for a total y displacement
  (yf-yo)
• At the top of the arc, vf0
• Let vo initial velocity, in positive direction,
  11.31 m/s
• Let g9.8 m/s2 , acceleration due to gravity, in
  negative direction.

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Part B) Make some definitions

• What it is asking for is how high the can flies
  above the release point it must be greater than
  4 m
• So we are solving for a total y displacement
  (yf-yo)
• At the top of the arc, vf0
• Let vo initial velocity, in positive direction,
  11.31 m/s
• Let g9.8 m/s2 , acceleration due to gravity, in
  negative direction.

Yes, he can catch it