1. 9/4 Acceleration  Text sections 2.1-3 and 1.5-6  HW “9/4 Airplane” due Friday 9/6 On web or in 213 Witmer for copying  For Thursday, look at text sections 2.7 and 3.1-2 Graphing and 2-D Motion  Suggested Problems: 2-25, 26, 29, 30

2. 0m/s Example Problem A block slides from rest down a ramp, across a level section, then down another ramp of equal slope. Ignore friction. On the lever section the block moves with a constant velocity of 4m/s. v = 4m/s left What is the block’s average velocity on the upper ramp? v f = v i = 4m/s The average of 0 and 4 is 2. v ave = 2m/s down the ramp

3. Average Velocity Average velocity is the “middle” velocity as well as  x/  t. Example: An object slows down from 35m/s to 5m/s, what is the average velocity? It took 6s to slow down, how far did the object move? What is its speed at 3s, the “mid-time?”

4. Acceleration  A ball rolls up and down a ramp as shown in the strobe photograph. Which way does the acceleration point or does the acceleration = 0? Turnaround point Ball rolling up the ramp v Pick a time interval, t i - t f and draw velocity vectors titi tftf Draw velocity vectors tail to tail Draw  v, (from i to f) which points the same direction as a. vfvf vivi vivi vfvf a points down the ramp. a = vv tt

5. Acceleration and Velocity Example: An object moving left slows down from 35m/s to 5m/s, what is the average velocity direction? It took 6s to slow down, what is the object’s acceleration, magnitude and direction? (Always think about  v.)  v = 30m/s to the right a = 5m/s 2 to the right

6. Acceleration at turnaround  A ball rolls up and down a ramp as shown in the strobe photograph. At the turnaround point, which way does the acceleration point or does the acceleration = 0 there? Turnaround point

7. Acceleration at turnaround Turnaround point Ball rolling up the ramp v Pick a time interval, t i - t f and draw velocity vectors tftf Copy velocity vectors tail to tail vivi vivi Turnaround point Ball rolling down the ramp titi vfvf v vfvf Draw  v, (from i to f) which points the same direction as a. Even though v = 0, v is still changing and there is acceleration!!!!

8. Acceleration a = vv tt is an “operational definition” in that it defines a procedure for finding and using a. Finding acceleration Using Acceleration

9. “Change in Velocity” Vector,  v  v = -4m/s left v = 8m/s v = 4m/s v = 0m/s v = -4m/s v = -8m/s v = -12m/s  v = -4m/s left The “change in velocity” vector may point with or against the velocity vector. Even though the object slows down, turns around, and speeds up in the opposite direction;  v is constant!

10. Acceleration a = vv tt  v = -4m/s left v = 8m/s v = 4m/s v = 0m/s v = -4m/s v = -8m/s v = -12m/s  v = -4m/s left v and a point opposite, slowing down v and a point the same direction, speeding up Acceleration is a vector that points in the same direction as the “change in velocity” vector. In this case, a = 4m/s/s left. In concept, it is “the amount and direction the velocity changes each second.”

11. Concepts so far- Displacement,  x (distance moved) Instantaneous Velocity, v (at a particular time) Average Velocity, v ave (average over time) Change in Velocity,  v (speeding up or slowing down) Acceleration, a (how much the velocity changes each second)

12. Problem: An object goes from a velocity of 15 m/s right to 6 m/s right in 3 seconds. Find the acceleration, both its size (magnitude) and its direction, (left or right). How do the directions of the velocity and acceleration compare? What is the object doing during these 3 seconds? How far did the object travel during these three seconds? Hint: What is the average velocity? What will the objects velocity be in three more seconds if the acceleration stays the same?

13. Problem: A bullet exits a rifle at 85m/s. The barrel is 0.75m long. What is the acceleration of the bullet? Don’t use text equations, just the relationships between displacement, time, velocity and acceleration

14. Finding acceleration v i = 10m/s v f = 40m/s  v = 30m/s right Return a = vv tt  t = 6s = 30 6 = 5m/s/s right

15. Problem:  A bear is running 4 m/s north. The acceleration of the bear is 3m/s 2 north. What is the bear’s velocity 2 seconds later? v = 10 m/s north  What is the bear’s average velocity? How far did the bear run during this time? v ave = 7 m/s north  x = 14 m north Return