unit 2 class notes
DESCRIPTION
The Kinematics Equations (1D Equations of Motion). Unit 2 Class Notes. Honors Physics. Day 1. Introduction to the 1D equations of motion. ACCELERATION. Leonardo. Michaelangelo. Raphael. Donatello. Derivations of the equations (a.k.a. where the turtles come from). Raphael. - PowerPoint PPT PresentationTRANSCRIPT
Unit 2 Class Notes
Honors Physics
The Kinematics Equations (1D Equations of Motion)
Day 1
Introduction to the 1D equations of motion
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Δx = x2 − x1
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v = ΔxΔt
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a = ΔvΔt
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Δv = v2 − v1
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s = d Δt
Δx 12 at
2 v1t
v22 v1
2 2aΔx
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v2 = v1 + at
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Δx = 12 t(v1 + v2)
ACCELERATION
Leonardo
Δx 12 at
2 v1t
Michaelangelo
v22 v1
2 2aΔx
v22
Raphael
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v2 = v1 + at
Donatello
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Δx = 12 t(v1 + v2)
Derivations of the equations (a.k.a. where the turtles come from)
Raphael …. slope of the line =
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a =(v2 − v1)
Δt
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v2 = v1 + at
Donatello
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Δx
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Δx = 12 Δt(v1 + v2)
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area = 12 h(b1 +b2)
…. area under curve =
What about Leonardo and Michelangelo?
Fighting with the turtles
List what you have (using “COMPATIBLE” units & using proper SIGNS)
Choose your warrior (in other words, “choose your equation”)
Solve the equation
Make sure your answer makes sense (both in MAGNITUDE & DIRECTION)
Make a decision as to which direction is POSITIVE & which is NEGATIVE
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v1 =100 kmh = 27.7 m
s
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v2 = 0
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t = 4.5sec
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v2 = v1 + at
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0 = 27.7777777 + a(4.5)
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a = −6.17 ms2
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Δx = 12 t(v1 + v2)
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Δx = 12 (4.5)(27.77777 + 0)
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Δx = 62.5m
2.
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v1 = 8 ms
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a = 2 ms2
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Δx =100m
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Δx = 12 at
2 + v1t
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100 = 12 (2)t 2 + 8t
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0 = t 2 + 8t −100
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t = −14.77sec, 6.77sec
3.
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v1 = 40 ms
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v2 = 20 ms
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Δx = 50m
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v22 = v1
2 + 2aΔx
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Δx = 12 t(v1 + v2)
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50 = 12 t(40 + 20)
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t =1.67sec
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202 = 402 + 2a(50)
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a = −12 ms2
4.
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v1 = 0 ms
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v2 = 60mph = 26.82 ms
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t = 3.5sec
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v2 = v1 + at
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26.82 = 0 + a(3.5)
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a = 7.66 ms2
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Δx0−1 = 12 at
2 + v1t
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=12 (7.66)(1)2 + 0(1)
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=3.83m
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Δx0−2 = 12 at
2 + v1t
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=12 (7.66)(2)2 + 0(2)
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=15.32m
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d0−1 = 3.83m d1−2 =15.32m − 3.83m =11.49m
5.
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v22 = v1
2 + 2aΔx
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v1 = v
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v2 = 0
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Δx = x€
02 = v 2 + 2ax
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a = −v 2
2x
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v22 = v1
2 + 2aΔx
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02 = (2v)2 + 2−v 2
2x
⎛
⎝ ⎜
⎞
⎠ ⎟Δx
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v1 = 2v
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v2 = 0
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Δx = ?
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a = −v 2
2x
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0 = 4v 2 −v 2
xΔx
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4v 2 =v 2
xΔx
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Δx = 4x
6.
1 2 3
1-2 2-3 1-3
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v1 = 0
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Δx23 = 20m
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t23 =1.40sec
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a = 3 ms2
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a = 3 ms2
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v1 = 0
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a = 3 ms2
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v2 =12.186 ms
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v3 =16.386 ms
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v2 =12.186 ms
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Δx = 24.75m
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t = 5.462sec
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v3 =16.386 ms
The 5th ninja?
Chris Farley
d rt
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Δx = 12 at
2 + v1t
TONIGHTS HWComplete pp. 61-62, #’s 61-69 (skip 62 & 66), 71, 72