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Equations of Motion!

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

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Equations of Motion!

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  1. Equations of Motion! Motion with Constant Velocity xf = x0 + vt Motion with Constant Acceleration vf = v0 + at xf = x0 +v0t + ½at2 vf2 = v02 + 2aΔx t a = 0 m/s2 The “big three” #2 can also be written as Δx = v0t + ½at2, since Δx = xf – x0

  2. 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!)

  3. Choose an Origin and Positive Direction Easiest choice! y0 = 0 m a = 9.8 m/s2 Δy will be positive

  4. 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 = v0t + ½at2 where Δy is half the height of the cliff and t = 3 s Initial: Top of cliff (v0 = 0) Final: Halfway down This gives Δy = ½ * (9.8 m/s2) * (9 s2) = 44.1 m Therefore, the cliff is 88.2 m high.

  5. How long does it take to fall the second half? Now, we need to use Δy = v0t + ½at2, but with different initial and final conditions. Initial: Halfway down (the rock will have a nonzero v0) 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 vf = v0 + at, and we get vhalfway = 9.8 m/s2 * 3s = 29.4 m/s

  6. Finishing the job Δy = 44.1 m v0 = 29.4 m/s a = 9.8 m/s2 Δy = v0t + ½at2 44.1 = (29.4)t + 4.9 t2

  7. 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.

  8. 44.1 = (29.4)t + 4.9 t2 Rearrange to get all terms on one side 4.9 t2 + 29.4 t – 44.1 = 0 a = 4.9 b =29.4 c = -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?

  9. 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!

  10. 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

  11. 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

  12. Meet the Position vs Time Graph

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

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

  15. 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

  16. 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

  17. 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

  18. 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

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

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

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