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Acceleration

Acceleration. Changing Motion. Objects with changing velocities cover different distances in equal time intervals. Acceleration. Rate at which an object’s velocity changes. When the velocity changes at a constant rate, the object has constant acceleration.

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Acceleration

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  1. Acceleration

  2. Changing Motion • Objects with changing velocities cover different distances in equal time intervals.

  3. Acceleration • Rate at which an object’s velocity changes. • When the velocity changes at a constant rate, the object has constant acceleration. • Average Acceleration = change in velocity during a time interval • Instantaneous acceleration = acceleration at any instant.

  4. Velocity-Time Graph • A velocity-time graph can be used to find acceleration • The slope of the velocity-time graph gives the acceleration. a = v / t a = (vf - vi) / (tf - ti )

  5. Positive and Negative Acceleration • Positive acceleration means that the acceleration vector is in the positive direction • Negative acceleration means that the acceleration vector is in the negative direction. • Positive and negative acceleration do not indicate speeding up or slowing down.

  6. Given:vi = 0 vf = 9.00 m/s t = 5.00 s Find:a = ? a = (vf - vi) / t = (9.00 m/s - 0)/(5.00 s) = 1.80 m/s2 Ex. #1: A shuttle bus starts from rest and reaches a velocity of 9.00 m/s in 5.00 s. Find the average acceleration of the bus.

  7. Given:vi = 0 m/s vf = 27 m/s a = 15 m/s2 Find:t = ? a = (vf - vi) / t t = (vf - vi) / a = (27 m/s - 0 m/s) / (15 m/s2) = 1.8 s Ex. #2: A motorcycle has an average acceleration of 15 m/s2. How much time is required for it to reach 27 m/s from rest?

  8. Velocity-Time Graph • The area between the velocity-time curve and the x-axis gives the displacement of the object. • Add all areas to get total displacement. • Negative areas indicate negative displacement

  9. Motion with Constant Acceleration • Equations can be derived from a velocity-time graph • vf = vi + aΔt(slope of v-t graph) • df = di + vi Δt + ½ aΔt2(area under curve-triangle + rectangle) • Δd = ½ (vi + vf) Δt(area under curve-trapezoid) • vf2 = vi2 + 2a Δd (combine 1st and 2nd equns to remove time)

  10. Given: t = 2.0 s vi = 16.2 m/s vf= 1.7 m/s Find: a = ? vf= vi + a t (vf – vi) / t = a (1.7 m/s -16.2 m/s) / 2.0 s = a = - 7.3 m/s2 Ex. 1: A boat on the Log Flume at Six Flags takes 2.0 s to slow down from 16.2 m/s to 1.7 m/s. What is the rate of acceleration of the boat?

  11. Given: t = 2.0 s vi = 16.2 m/s vf = 1.7 m/s Find: d = ? d = ½ (vi + vf) t = ½ (16.2 m/s + 1.7 m/s)(2.0 s) = 18 m Ex. 2: How far does the same boat on the Log Flume travel while slowing down?

  12. Given: vi = 0 m/s vf = 16.2 m/s a = 5.2 m/s2 Find: d = ? vf2 = vi2 + 2ad (vf2 – vi2) / 2a = d [(16.2 m/s)2 - (0)2] /(2*5.2 m/s2) = d = 25 m Ex. 3: Back on the Log Flume, the next boat accelerates down the slide at 5.2 m/s2. If the boat starts from rest, how long is the slide?

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