Download Presentation
## Announcements

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Announcements**• No class nextMonday (MLK day)**Equations of Motion**Tractable cases §2.5–2.6**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?**distance units**Area under a v-t graph area = (a m/s)(b s) = ab m speed (m/s) a b time (s)**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**Visualize Acceleration**Young and Freedman, Fig. 2.8 Board Work: • Signs of v • Signs of a**slope = velocity**d t slope = acceleration v area = distance t a area = velocity t Acceleration Starting 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?