1 / 8

Data Accurate Measurements

Data Accurate Measurements. The Protractor. Tycho Brahe 1546 –1601 Accuracy to one minute of arc 1/60 degree = 1 nautical mile sextant. Aristotle 384 – 322 BC “Heavier things fall faster” Galileo Galilei 1564 –1642 “All objects fall with the same constant acceleration.”

Download Presentation

Data Accurate Measurements

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. DataAccurate Measurements

  2. The Protractor • Tycho Brahe 1546 –1601 • Accuracy to one minute of arc • 1/60 degree • = 1 nautical mile • sextant

  3. Aristotle 384 – 322 BC • “Heavier things fall faster” • Galileo Galilei 1564 –1642 • “All objects fall with the same constant acceleration.” • The change in velocity is proportional to the elapsed time. • Δv ~ Δt • Δv= g∙Δt

  4. Acceleration • Acceleration – How quickly the velocity changes • a = Δv / t = (v2-v1)/t • Velocity – How quickly the position changesv = s/t = Δx/t = (x2-x1)/t • When things fall acceleration = a = 9.8 m/sec/sec ≈ 10 m /sec2

  5. Formulas • ∆ v = g ∆t change in v ~ time • v2 − v1 = g (t2 − t1) (time when) • v2 = v1 + gt (elapsed time) • vavg= v̄ = ½ (v1 + v2) • distance = s = ∆x = x2 − x1 = v̄ t • s= ½ (v1 + v2) t • s = ½ (v1 + v1 + gt) t • s = v1 t + ½ gt2

  6. More Formulas • Algebra Review: (x + y )∙(x – y ) f o i l • = x² – xy + xy – y² • a = (v2 – v1)/t • s = ½(v2+v1)∙t (average velocity)∙ time • Multiply : • as = ½(v2² –v1²) •  2as = v2² –v1²

  7. Prob Set 4: 9 • An arrow is shot straight up with a speed of 40 meters per second. Three seconds later it hits a bird. • How fast was the arrow going when it hit the bird ? • g = −10 m/s2 • t = 3 s • v0 = 40 m/s • v1 = v0 + a⋅t = 40m/s – 10m/s2 ⋅ 3s = • 40m/s – 30m/s = 10 m/s • How high was the bird ? • s = v̅ t = ½ (40+10) ⋅t = 25m/s ⋅ 3 s = 75 m

  8. Suppose the Distance is Given • The arrow is shot with a speed of 40 meters per second. The bird is 50 meters up. • How fast was the arrow going when it hit the bird ? • g = −10 m/s2 s = 50 m t = ? • v1 = 40 m/s • v2 = v1 + a ⋅ t Won’t work • 2as = v2² – v1²  v2 is unknown • v2² =v1² + 2as = 40² + 2 ⋅ (–10)⋅ 50 • = 1600 – 1000 = 600 • v2 = √600 = 24.5 m/s up or 24.5 m/s down

More Related