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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?

Area under a v-t graph speed (m/s) time (s) area = (a m/s)  (b s) = ab m a b distance units

Constant-Velocity Motion v =  x/  t = constant throughout process  x = v  t x f = x i +  x = x i + v  t 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 =  v/  t  v = velocity change over time  t  v = a  t v = v 0 +  v = v 0 + a  t

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: 2.Signs of v 3.Signs of a

Acceleration Starting from a traffic light that turns green d t v t a t area = velocity area = distance slope = velocity slope = acceleration

Equations of Motion What are velocity and position under conditions of constant acceleration?

Formulas from Constant x-Acceleration Velocity change  v = a  t Velocity v t = v 0 +  v = v 0 + a  t Position change  x = v 0  t + 1/2 a (  t) 2 Position x t = x 0 + v 0  t + 1/2 a (  t) 2

Another Form (constant a) If you don’t know  t and want v: x = x 0 + v 0  t + 1/2 a (  t) 2  t =  v/a x – x 0 = v 0  v/a + 1/2 a (  v/a) 2 2a (x–x 0 ) = 2v 0 (v–v 0 ) + (v–v 0 ) 2 2a (x–x 0 ) = 2vv 0 – 2v v 2 – 2vv 0 + v 0 2 2a (x–x 0 ) = 2vv 0 – 2vv 0 + v 2 + v 0 2 – 2v 0 2 2a (x–x 0 ) = v 2 – v 0 2 v 2 = v a (x–x 0 ) Do units work?

Another Form (constant a) If you don’t know a but know v, v 0, and  t: x = x 0 + v 0  t + 1/2 a (  t) 2 a =  v/  t = (v–v 0 )/  t x = x 0 + v 0  t + 1/2 ( (v–v 0 )/  t ) (  t) 2 x – x 0 = v 0  t + 1/2 v  t – 1/2 v 0  t x – x 0 = v 0  t – 1/2 v 0  t + 1/2 v  t x – x 0 = 1/2 (v 0 + v)  t 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/s 2 and if the car steps on the gas the car will accelerate at 2.3 m/s 2. 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?