unit 2 class notes

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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 Presentation

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Unit 2 Class Notes

Honors Physics

The Kinematics Equations (1D Equations of Motion)

Day 1

Introduction to the 1D equations of motion

Δx = x2 − x1

v = ΔxΔt

a = ΔvΔt

Δv = v2 − v1

s = d Δt

Δx 12 at

2 v1t

v22 v1

2 2aΔx

v2 = v1 + at

Δx = 12 t(v1 + v2)

ACCELERATION

Leonardo

Δx 12 at

2 v1t

Michaelangelo

v22 v1

2 2aΔx

v22

Raphael

v2 = v1 + at

Donatello

Δx = 12 t(v1 + v2)

Derivations of the equations (a.k.a. where the turtles come from)

Raphael …. slope of the line =

a =(v2 − v1)

Δt

v2 = v1 + at

Donatello

Δx

Δx = 12 Δt(v1 + v2)

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

v1 =100 kmh = 27.7 m

s

v2 = 0

t = 4.5sec

v2 = v1 + at

0 = 27.7777777 + a(4.5)

a = −6.17 ms2

Δx = 12 t(v1 + v2)

Δx = 12 (4.5)(27.77777 + 0)

Δx = 62.5m

2.

v1 = 8 ms

a = 2 ms2

Δx =100m

Δx = 12 at

2 + v1t

100 = 12 (2)t 2 + 8t

0 = t 2 + 8t −100

t = −14.77sec, 6.77sec

3.

v1 = 40 ms

v2 = 20 ms

Δx = 50m

v22 = v1

2 + 2aΔx

Δx = 12 t(v1 + v2)

50 = 12 t(40 + 20)

t =1.67sec

202 = 402 + 2a(50)

a = −12 ms2

4.

v1 = 0 ms

v2 = 60mph = 26.82 ms

t = 3.5sec

v2 = v1 + at

26.82 = 0 + a(3.5)

a = 7.66 ms2

Δx0−1 = 12 at

2 + v1t

=12 (7.66)(1)2 + 0(1)

=3.83m

Δx0−2 = 12 at

2 + v1t

=12 (7.66)(2)2 + 0(2)

=15.32m

d0−1 = 3.83m d1−2 =15.32m − 3.83m =11.49m

5.

v22 = v1

2 + 2aΔx

v1 = v

v2 = 0

Δx = x€

02 = v 2 + 2ax

a = −v 2

2x

v22 = v1

2 + 2aΔx

02 = (2v)2 + 2−v 2

2x

⎝ ⎜

⎠ ⎟Δx

v1 = 2v

v2 = 0

Δx = ?

a = −v 2

2x

0 = 4v 2 −v 2

xΔx

4v 2 =v 2

xΔx

Δx = 4x

6.

1 2 3

1-2 2-3 1-3

v1 = 0

Δx23 = 20m

t23 =1.40sec

a = 3 ms2

a = 3 ms2

v1 = 0

a = 3 ms2

v2 =12.186 ms

v3 =16.386 ms

v2 =12.186 ms

Δx = 24.75m

t = 5.462sec

v3 =16.386 ms

The 5th ninja?

Chris Farley

d rt

Δx = 12 at

2 + v1t

TONIGHTS HWComplete pp. 61-62, #’s 61-69 (skip 62 & 66), 71, 72

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