Chapter 2 Forces Acceleration; Friction; Gravity; Momentum; Newton’s 1st, 2nd, and 3rd Laws.
1.6 - Acceleration due to Gravity - Clarke's...
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1.6 Acceleration due to Gravity 1 February 11, 2016 Equation ( ) ∆d = v f + v i 2 ∆t Equation 1 Equation 2 Equation 3 Equation 4 Equation 5 v f = v i + a av ∆t v f 2 = v i 2 +2a av ∆d ∆d= v i ∆t + 1 2 a av ∆t 2 Δd=v f Δt 1 2 a av ∆t 2 Five Key Equations for Motion with Uniform Acceleration Section 1.5 In a uniform acceleration problem, you are given v f , v i , and a av . You need to solve for ∆t. Drag and drop the checkmark to show which equation(s) you can use to solve for ∆t. Check Answer
Transcript of 1.6 - Acceleration due to Gravity - Clarke's...
1.6 Acceleration due to Gravity
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Equation
( )∆d =vf + vi2
∆tEquation 1
Equation 2
Equation 3
Equation 4
Equation 5
vf = vi + aav∆t
vf2 = vi2 + 2aav∆d
∆d = vi∆t + 12
aav∆t2
Δd = vf Δt 12 aav∆t2
Five Key Equations for Motion with Uniform AccelerationSection 1.5
In a uniform acceleration problem, you are given vf , vi , and aav . You need to solve for ∆t. Drag and drop the checkmark to show which equation(s) you can use to solve for ∆t.
CheckAnswer
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Activity Acceleration due to Gravity
Purpose:
To see which object will have a quicker acceleration rate when dropped and allowed to freefall: a shuttlecock, a wiffleball or a baseball.
Time (s) Position Time (s) Velocity
Acceleration Rate:
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Uniform Vertical AccelerationSection 1.6
For your physics class, you drop a ball from a tall building. Complete the table below on a sheet of paper by calculating the distances the ball falls for the first 5 s. Then watch the animation to check your answers.
CheckAnswer
∆d = vi∆t +12 aav∆t2 (m)∆t (s)
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Discussion
Consider the distance the ball falls. What pattern do you see? Explain the pattern.
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ANSWERS
TEACHER'S NOTES LESSON
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Five Key Equations for Motion with Uniform AccelerationSection 1.5
In a uniform acceleration problem, you are given vf , vi , and aav . You need to solve for ∆t. Drag and drop the checkmark to show which equation(s) you can use to solve for ∆t.
Back
Equation
Equation 1
Equation 2
Equation 3
Equation 4
Equation 5
vf = vi + aav∆t
vf2 = vi2 + 2aav∆d
∆d = vi∆t + 12
aav∆t2
( )∆d =vf + vi2
∆t
Δd = vf Δt 12 aav∆t2
1.6 Acceleration due to Gravity
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February 11, 2016
Uniform Vertical AccelerationSection 1.6
For your physics class, you drop a ball from a tall building. Complete the table below on a sheet of paper by calculating the distances the ball falls for the first 5 s. Then watch the animation to check your answers.
= (9.8 m/s2) (0 s)2 = 0
= (9.8 m/s2) (1 s)2 = 4.9
= (9.8 m/s2) (2 s)2 = 19.6
= (9.8 m/s2) (3 s)2 = 44.1
= (9.8 m/s2) (4 s)2 = 78.4
= (9.8 m/s2) (5 s)2 = 122.5
121212121212
∆d = vi∆t +12 aav∆t2 (m)∆t (s)
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0
Back
Discussion
Consider the distance the ball falls. What pattern do you see? Explain the pattern.
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Answers for Discussion Questions:
Slide 9
For every equal interval of time, the distance the ball falls increases. As the ball falls toward Earth, its speed is increasing. Since the distance travelled increases for each successive increment of time, then the ball must be gaining speed (accelerating) as it falls.
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