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## PowerPoint Slideshow about ' Announcements' - dulcea

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Announcements

- No class nextMonday (MLK day)

Find Position from Velocity

- Generally: velocity is slope of a position-time graph.
- Conversely, position is the area under a velocity-time graph.
- What is this when v is constant?

Constant-Velocity Motion

- v = Dx/Dt = constant throughout process
- Dx = vDt
- xf = xi + Dx = xi + vDt
- Can also use this with average v

Find Velocity from Acceleration

- General case: acceleration is slope of a velocity-time graph.
- Conversely, velocity is the area under an acceleration-time graph.
- What is this when a is constant?

Constant-Acceleration Motion

- Instantaneous accel = average accel
- a = Dv/Dt
- Dv = velocity change over time Dt
- Dv = aDt
- v = v0 + Dv = v0 + aDt

Acceleration on an x-t Graph

- Velocity is the slope of a position-time graph
- Acceleration means a changing slope
- A constant slope means a straight x-t line
- A varying slope means a curved x-t line

- Positive acceleration = concave up

- Negative acceleration = concave down

slope = velocity

d

t

slope = acceleration

v

area = distance

t

a

area = velocity

t

AccelerationStarting from a traffic light that turns green

Equations of Motion

- What are velocity and position under conditions of constant acceleration?

Formulas from Constant x-Acceleration

- Velocity change Dv = aDt
- Velocity vt = v0 + Dv = v0 + aDt
- Position change Dx = v0Dt + 1/2 a (Dt)2
- Position xt = x0 + v0Dt + 1/2 a (Dt)2

Another Form (constant a)

- If you don’t know Dt and want v:

x = x0 + v0Dt + 1/2a (Dt)2Dt = Dv/a

x – x0 = v0 Dv/a + 1/2a (Dv/a)2

2a (x–x0) = 2v0 (v–v0) + (v–v0)2

2a (x–x0) = 2vv0 – 2v02 + v2 – 2vv0 + v02

2a (x–x0) = 2vv0 – 2vv0 + v2 + v02 – 2v02

2a (x–x0) = v2 – v02

v2 = v02 + 2a (x–x0)

Do units work?

Another Form (constant a)

- If you don’t know a but know v, v0, and Dt:

x = x0 + v0Dt + 1/2a (Dt)2a = Dv/Dt = (v–v0)/Dt

x = x0 + v0 Dt + 1/2((v–v0)/Dt) (Dt)2

x – x0 = v0 Dt + 1/2v Dt – 1/2v0 Dt

x – x0 = v0 Dt – 1/2v0 Dt + 1/2v Dt

x – x0 = 1/2 (v0 + v)Dt

Do units work?

Example Problem

A car 3.5 m in length traveling at 20 m/s approaches an intersection. The width of the intersection is 20 m. 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 and if the car steps on the gas the car will accelerate at 2.3 m/s2. The light will be yellow for 3 s.

To avoid being in the intersection when the light turns red, should the driver use the brake or the gas?

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