1. PHSX 114, Wednesday August 27 Reading for this lecture: Chapter 2 (2-5 thru 2-7)Reading for this lecture: Chapter 2 (2-5 thru 2-7) Reading for the next lecture: Chapter 2 (2.8), Chapter 3 (3-1 thru 3-4)Reading for the next lecture: Chapter 2 (2.8), Chapter 3 (3-1 thru 3-4) Homework for this lecture: Chapter 2, question: 15; problems: 20, 23, 27, 33, 34.Homework for this lecture: Chapter 2, question: 15; problems: 20, 23, 27, 33, 34.

2. Terms for describing motion Position xPosition x Displacement Δx=x 2 - x 1Displacement Δx=x 2 - x 1 velocity = displacement/time interval = Δx/Δtvelocity = displacement/time interval = Δx/Δt acceleration = change in velocity/time interval = Δv/Δtacceleration = change in velocity/time interval = Δv/Δt change in acceleration has no (commonly used) special namechange in acceleration has no (commonly used) special name

3. Acceleration is the rate of change of the velocity Average acceleration = change in velocity/time interval = Δv/ΔtAverage acceleration = change in velocity/time interval = Δv/Δt Instantaneous acceleration is the limit of the average acceleration as Δt approaches zeroInstantaneous acceleration is the limit of the average acceleration as Δt approaches zero For those who know calculus: If v(t) gives the velocity as a function of time, then the instantaneous acceleration, a(t), is the derivative of v(t) with respect to t.For those who know calculus: If v(t) gives the velocity as a function of time, then the instantaneous acceleration, a(t), is the derivative of v(t) with respect to t. ExamplesExamples

4. Your Turn: A runner starts from rest and accelerates at a=6 m/s 2. What is her speed after 5 s? Answer: 30 m/s (Δv=v 2 -v 1 =aΔt=(6 m/s 2 )(5s)=30 m/s; v 1 =0, so v 2 =30 m/s.)

5. What does the sign of the acceleration (positive/negative) signify? Answer: the sign of Δv (not direction of motion)Answer: the sign of Δv (not direction of motion) Example of positive acceleration: v 1 =10 m/s, v 2 =20 m/s, Δt = 5 s, gives a=(20-10)/5 = 2 m/s 2. Note: speeding up with velocity in positive directionExample of positive acceleration: v 1 =10 m/s, v 2 =20 m/s, Δt = 5 s, gives a=(20-10)/5 = 2 m/s 2. Note: speeding up with velocity in positive direction Another example of positive acceleration: v 1 =-20 m/s, v 2 =-10 m/s, Δt = 5 s, gives a=(-10-(-20))/5 = 2 m/s 2. Note: slowing down with velocity in negative directionAnother example of positive acceleration: v 1 =-20 m/s, v 2 =-10 m/s, Δt = 5 s, gives a=(-10-(-20))/5 = 2 m/s 2. Note: slowing down with velocity in negative direction

6. Your Turn: a) If my acceleration is negative and I'm moving in the positive direction, am I speeding up or slowing down? b) If my acceleration is negative and I'm moving in the negative direction, am I speeding up or slowing down? Answer: a) slowing down (example: v 1 =20 m/s, v 2 =10 m/s, Δt = 5 s, gives a=(10-20)/5 = -2 m/s 2 b) speeding up

7. Motion with constant acceleration A set of equations describes the position x and velocity v as a function of timeA set of equations describes the position x and velocity v as a function of time Notation: x 0 is the position at time t=0, v 0 is the velocity at time t=0,Notation: x 0 is the position at time t=0, v 0 is the velocity at time t=0, If acceleration is constant -- v ave =½ (v+ v 0 )If acceleration is constant -- v ave =½ (v+ v 0 )

8. Equations for constant acceleration v = v 0 + atv = v 0 + at derivation: a=(v- v 0 )/t  at=(v- v 0 )  v = v 0 + atderivation: a=(v- v 0 )/t  at=(v- v 0 )  v = v 0 + at x = x 0 + v 0 t + ½at 2x = x 0 + v 0 t + ½at 2 derivationderivation v 2 = v 0 2 +2a(x-x 0 )v 2 = v 0 2 +2a(x-x 0 ) exampleexample

9. Strategy for kinematics problems 1. Make a sketch 2. Write down knowns and unknowns 3. Find an expression that relates knowns to unknowns 4. Solve for your unknowns 5. Sanity checks: are units OK? Is the number reasonable?

10. Your turn: At t=0 a car has velocity 10 m/s and a constant acceleration of 1 m/s 2. How far does the car travel in 10 s? Answer: 150 m x = x 0 + v 0 t + ½at 2 ; x 0 =0, v 0 =10 m/s, a=1 m/s 2, t=10 s

11. Application of Constant Acceleration -- Falling objects All objects fall under gravity with the same acceleration.All objects fall under gravity with the same acceleration. Our first real physics statementOur first real physics statement Let's test it!Let's test it!

12. Acceleration due to gravity -- g = 9.80 m/s 2. Value varies by a few percent depending on the location on the EarthValue varies by a few percent depending on the location on the Earth Example problemsExample problems