1 / 13

Physics Chp 2

Physics Chp 2. Displacement – is the difference between the intitial position and the final position. It does not matter how far you really went. ex. Flying to Boston via Sea or Salt Lake or Houston. All 3 you have the same displacement and it is symbolized as Δ x .

Download Presentation

Physics Chp 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Physics Chp 2

  2. Displacement – is the difference between the intitial position and the final position. It does not matter how far you really went. ex. Flying to Boston via Sea or Salt Lake or Houston. All 3 you have the same displacement and it is symbolized as Δx. Needs a value and direction with units

  3. Velocity is the change in x divided by time v = Δx/ Δt Avg. velocity is just comparing any two points of data. Instantaneous velocity is at one instant in time and is the tangent to the graph of Δx vs Δt

  4. Slope of tangent to the graph gives the velocity at that instant in time.

  5. Acceleration a = Δv/ Δt graphical const vs variable

  6. Equations when a = constant vf = vo +aΔt Δx = 1/2(vo + vf)Δt Δx = voΔt + 1/2a(Δt)2 vf2 = vo2 + 2aΔx

  7. Ex Find the final velocity if you toss a ball from your car traveling North at 200 m/s and your arm accelerates it 50 m/s2 N for 0.3 s.

  8. Vo = 200 m/s N a = 50 m/s2 N t = 0.3 s • vf = vo +aΔt • vf = 200 m/s + 50 m/s2(0.3s) • Vf =

  9. Free Fall is just when a = -9.81 m/s2 Ex If it takes 13 s for a rock dropped off a cliff to hit the bottom, how high is the cliff? Ignore the echo issue.

  10. vo = 0 m/s a = -9.81 m/s2 t = 13s • Δx = voΔt + 1/2a(Δt)2 • Δx = 0m/s(13s) + ½(-9.81m/s2)(13s)2 • Δx = -830m cliff is really 830m Why

  11. Ex How fast did the rock hit the bottom of the cliff?

  12. vo = 0 a = -9.81 m/s2 t = 13s ∆x = -830m • vf2 = vo2 + 2aΔx • vf2 = 02 + 2(-9.81 m/s2)(-830m) • vf2 = • vf = ?????

More Related