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Rectilinear Motion Continued

Rectilinear Motion Continued. Sections 4.2 and 4.3. Updates & Reminders. The date and time of the final: Thursday, December 11 from 10 a.m. to 12 p.m. MLT 208. There is no lab this week. Test #1 will take place on Thursday, 9/18.

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Rectilinear Motion Continued

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  1. Rectilinear Motion Continued Sections 4.2 and 4.3

  2. Updates & Reminders • The date and time of the final: Thursday, December 11 from 10 a.m. to 12 p.m. MLT 208. • There is no lab this week. • Test #1 will take place on Thursday, 9/18. • Test covers Chapters 1, 3, and 4 (only constant motion), Galileo, Knowledge & Faith, graphing) • There is no Mallard-based reading quiz prior to class on Thursday.

  3. The Position-Time graph (P-T) • Slope is velocity (+/- speed) • y-intercept is position at time = 0 • y = mx + b (algebraic relationship) • distance = velocity*time + distanceo • x = xo + vt (equation 1) • Solving practical problems…

  4. Sample Problems • An industrial sewing machine can stitch cloth at a rate of 3.8cm/s. During an 8-hour continuous run, how many METERS of stitching can be accomplished? • The current men's world record for the 100-meter dash is 9.58s. Usain Bolt from Jamaica set the record on August 16, 2009 at the World Championships in Berlin, Germany. What was Bolt’s average speed over this interval? • A lightning flash occurs and thunder is heard 2.4s later. How far away was the lightning bolt at its nearest point? Assume that the speed of sound is 343m/s. Be aware that the speed of light is so high that it can travel around Earth 7.5 times in one second. • In the above question, if the thunder continues to rumble for another 3.9s after the start, what was the most distant part of the lightning bolt? • A surface ship is bouncing sonar waves off the ocean bottom to determine depth. If a “ping” takes 1.534s to go down and return, how deep is the ocean beneath the ship? Assume that speed of the sonar wave is 1,473m/s.

  5. The Velocity-Time Graph • Slope = change in velocity / change in time which is acceleration • y-intercept equals velocity at time = 0 • y = mx + b • velocity = acceleration*time + velocityo • v = vo + at (equation 2) • Solving a practical problem…

  6. The V-T graph details • Constant (non-accelerated) motion • Slope = zero • Uniformly accelerated motion • Slope constant, but does not equal zero • Non-uniformly accelerated motion • Slope of tangent line at a given point gives accel. • One cannot determine initial position based on a V-T graph. • The area under a V-T graph is displacement.

  7. Uniform Acceleration V-T Graphs • The relationship between variables in uniformly accelerated motion. • x = xo +vot + ½ at2 • A practical example… • We will address this subject matter further in Chapter 6. • This week’s reflection will be based on a reading of the transitional Chapter 5.

  8. Kinematic Relationships • x = xo + vt (constant velocity) (equation 1) • v = vo + at (equation 2) • x = xo + vot + ½at2 (equation 3) • Substituting t from equation 2 into equation 3 results in 2aΔx = v2 – vo2 (equation 4) • Extra credit project for one point: • Demonstrate the derivation of equation 4. • Turn in written proof on Thursday at start of class.

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