3. a = 0 m/s 2 Motion with Constant Velocity x f = x 0 + vt Motion with Constant Acceleration v f = v 0 + at x f = x 0 + v 0 t + ½at 2 v f 2 = v 0 2 + 2aΔx Equations of Motion! #2 can also be written as Δx = v 0 t + ½at 2, since Δx = x f – x 0 t t The “big three”

4. AP Level Kinematics Problem A rock is dropped off a cliff and falls the first half of the distance to the ground in 3.0 seconds. How long will it take to fall the second half? Ignore air resistance. (Hint: Figure out the height of the cliff first!)

5. Choose an Origin and Positive Direction Easiest choice! y 0 = 0 m a = 9.8 m/s 2 Δy will be positive

6. How High is the Cliff? A rock is dropped off a cliff and falls the first half of the distance to the ground in 3.0 seconds. For the first half of the fall, we can use Δy = v 0 t + ½at 2 where Δy is half the height of the cliff and t = 3 s This gives Δy = ½ * (9.8 m/s 2 ) * (9 s 2 ) = 44.1 m Therefore, the cliff is 88.2 m high. Initial: Top of cliff (v 0 = 0) Final: Halfway down

7. How long does it take to fall the second half? Now, we need to use Δy = v 0 t + ½at 2, but with different initial and final conditions. Initial: Halfway down (the rock will have a nonzero v 0 ) Final: Just about to hit the ground. Δy = 44.1 m (second half is just as far as first half) To find the rock’s velocity halfway down, we must use v f = v 0 + at, and we get v halfway = 9.8 m/s 2 * 3s = 29.4 m/s

8. Finishing the job Δy = 44.1 m v 0 = 29.4 m/s a = 9.8 m/s 2 Δy = v 0 t + ½at 2 44.1 = (29.4)t + 4.9 t 2

9. Sometimes (semi-rarely), you will need to use the quadratic equation to calculate time when the object has an initial position, initial velocity and a nonzero acceleration. One of your homework problems uses it; make sure you also know the quadratic equation by heart. Remember to check if a, b or c are negative.

10. 44.1 = (29.4)t + 4.9 t 2 Rearrange to get all terms on one side 4.9 t 2 + 29.4 t – 44.1 = 0 a = 4.9b =29.4c = -44.1 Using quadratic, we get t = -7.24 s or 1.24 s The first half takes 3 seconds… The second half takes only 1.24 seconds! Why?

11. But WAIT!!!!!! There’s an easier, non-quadratic-ier way to solve it! Once you find that the cliff is 88.2 m tall, and you know that the first half of the drop takes 3 seconds… Find the total time of the whole drop, and subtract 3 seconds! This should get you the same result – try it!

12. Tips for Solving Kinematics Problems Read critically (“dropped”, “comes to stop”, “highest point reached”) Direction of a vector determines its sign. Record all of the given information Memorize the big three! You will not have a calculator on the AP Exam multiple choice, so if the calculations become too hard, go back and try a different approach

13. Final Tip If you encounter a toughie, try to step back, breathe and look at the scenario as a whole. There may be a shortcut or technique that makes the problem easier and saves you time! This can’t really be put into a series of steps, you just need to practice by studying hard and spending time on your homework

15. Meet the Position vs Time Graph

16. Interpreting x vs t graphs The slope of the tangent line represents the instantaneous velocity of the object at any one instant in time

17. The slope of a secant line represents the average velocity of the object between two moments in time. v av = Δx/Δt

18. Example from 2008 Exam The figure to the left shows the position vs. time graphs for two objects, A and B, that are moving along the same axis. The two objects have the same velocity (A) at t = 0 s. (B) at t = 2 s. (C) at t = 3 s. (D) at t = 4 s. (E) never

19. Example from 2008 Exam The figure to the left shows the position vs. time graphs for two objects, A and B, that are moving along the same axis. The two objects have the same velocity (A) at t = 0 s. (B) at t = 2 s. (C) at t = 3 s. (D) at t = 4 s. (E) never v v

20. Example from 2009 Exam The graph below represents position vs. time for a sprinter at the start of a race. Her average speed during the interval between 1 second and 2 seconds is most nearly: (A) 2 m/s (B) 4 m/s (C) 5 m/s (D) 6 m/s (E) 8 m/s

21. Example from 2009 Exam The graph below represents position vs. time for a sprinter at the start of a race. Her average speed during the interval between 1 second and 2 seconds is most nearly: (A) 2 m/s (B) 4 m/s (C) 5 m/s (D) 6 m/s (E) 8 m/s

22. However, x vs t graphs also have a SECRET PROPERTY hidden within them... …but I’m not going to tell you until tomorrow.

23. JUSTKIDDINGHEREWEGO The concavity of the x vs t graph tells you the acceleration of the object!