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Acceleration

Acceleration. To initiate, change or stop a movement Running Jumping Throwing Swinging Requires that our body, limbs, segments, joints must speed up and/or slow down. Acceleration.

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Acceleration

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  1. Acceleration To initiate, change or stop a movement Running Jumping Throwing Swinging Requires that our body, limbs, segments, joints must speed up and/or slow down.

  2. Acceleration • def: a change in velocity vector (MAGNITUDE and/or DIRECTION) over time. The ability to speed up or slow down • Positive acceleration (increase in velocity) • Negative acceleration (decrease in velocity) • DECELERATION - reduction in velocity

  3. Linear Acceleration • Linear acceleration (a) • a = ∆ in velocity/ ∆ in time = ∆v/∆t • UNITS Englishmetric ft/sec2 m/sec2 • if v = m/sec; t = sec • m/sec/sec = m/sec2

  4. Linear Acceleration - example • Acie Law dribbles at 15 ft/sec 15 feet from the basket moving in a straight line. When he reaches the basket 1.5 sec. later, Acie’s velocity is 26 ft/sec Find the acceleration!! • Given: v1 = 15 ft/sec ∆t = 1.5 sec • v2 = 26 ft/sec • Find: average linear acceleration (a)

  5. Linear Acceleration - example • DIAGRAM: • Formula:: a = ∆v/∆t • a = (v2 - v1)/∆t • a = 26 - (15 ft/sec)/1.5 sec • a = 11 ft/sec / 1.5 sec • a = 7.33 ft/sec2 V1 V2 26 ft/sec 15 ft/sec

  6. Linear Acceleration - example • A Cubs pitcher throws a 90 mph (132 ft/sec) fastball. Mark McGuire strikes the ball for his 61st homerun in 1998. The ball leaves very quickly in the opposite direction at 136 mph (200 ft/sec). The contact time of the bat on the ball is 0.01 sec. Find the acceleration!! • Given: v1 = -132 ft/sec ∆t = 0.01 sec • v2 = 200 ft/sec • Find: average linear acceleration (a)

  7. Linear Acceleration - example V1 • DIAGRAM: • Formula:: a = ∆v/∆t • a = (v2 - v1)/∆t • a = 200 -(-132 ft/sec) • a = 332 ft/sec / 0.01 sec • a = 33,200 ft/sec2 V2

  8. Acceleration – Changing Direction • Jessie is a striker in soccer. He is running at 8.6 m/sec. Jessie turns 37° and is now running at 7.9 m/sec. If the turn was accomplished in 1.8 sec, what is the acceleration? • Given: v1 = 8.6 m/sec v2 = 7.9 m/sec (Q) = 37° ∆t = 1.80 sec • Find: linear acceleration (a)

  9. turning and acceleration ∆v • DIAGRAM: • Formula: a = ∆v/∆t C2 = A2 + B2 - 2•A•B•cosQ Solution: • C2 = A2 + B2 - 2•A•B•cosq C = √(8.62 + 7.92 m/sec - 2•8.6•7.9• cos 37°) C = √(73.96 – 62.41 • (0.799)) C = √(24.12 m2/sec2) ∆v = C = 4.91 m/sec 8.6 m/sec 7.9 m/sec 37°

  10. turning and acceleration ∆v • DIAGRAM: • Solution: • ∆v = C = 4.91 m/sec a = ∆v/∆t a = 4.91 m/sec /1.80 sec a = 2.73 m/sec2 8.6 m/sec 7.9 m/sec 37°

  11. Angular Acceleration • angular acceleration (a) • a = ∆ in angular velocity/ ∆ in time = ∆w/∆t • UNITS Englishmetric rad/sec2 rad/sec2 • deg/sec2 deg/sec2

  12. Angular Acceleration • Randy Johnson is pitching a fastball at a speed of 103 mph. At 0.2 sec into his throw, the angular velocity of the left elbow is 260 °/sec. Two frames later, his elbow is extending at 1310°/sec. If the film speed is 30 frames/sec, what is the angular acceleration of the elbow joint?

  13. Angular Acceleration • Randy Johnson is pitching a fastball at a speed of 103 mph. At 0.2 sec into his throw, the angular velocity of the left elbow is 260 °/sec. Two frames later, his elbow is extending at 1310°/sec. If the film speed is 30 frames/sec, what is the angular acceleration of the elbow joint?

  14. Given: w1 = 260°/sec w2 = 1310°/sec • t1 = 0.20 sec film speed = 30 fr/sec • 2 frames elapsed • Find: a • DIAGRAM: A w2 w1

  15. Formula: a = ∆w/∆t • a = (w2 - w1)/(t 2 - t1) • ∆t = 1/30 sec/fr • 2 fr • ∆t = 0.067 sec • ∆ t = 0.067 sec • Solution: a = (w2 - w1)/(t 2 - t1) • a = (1310- 260 °/sec)/(0.267 - 0.20 sec) • a = (1050 °/sec)/(0.067 sec) • a = (15,749.99 °/sec2)

  16. Angular Acceleration • We film with a digital video camera a fast-pitch softball pitcher is throwing to home plate. The film speed is 30 frames/sec. Here, we’ll assume that the shoulder joint is the axis of rotation and the elbow is extended. Susan’s arm is 0.82 m long. Her arm is in an absolute angular position of 150° 6 frames after we start filming. On the next two consecutive frames, Susan’s arm is in a position of 178° and 225° respectively. Find the angular acceleration!

  17. Angular Acceleration • Given: q6 = 150° q7 = 178° q8 = 225° • r = 0.82 m t1 = 6/30 sec film speed = 30 fr/sec • Find: a q6 = 150° Diagram: origin q7 = 178° q8 = 225°

  18. Formula: a = ∆w/∆t • w1 = ∆q/∆t w2 = ∆q/∆t • w1 = (178-150°)/(1/30 sec) w2 = (225-178°)/(1/30 sec) • w1 = (28°)/(1/30 sec) w2 = (47°)/(1/30 sec) • w1 = (840 °/sec) w2 = (1410°/sec) q6 = 150° 0° q7 = 178° q8 = 225°

  19. Formula: a = ∆w/∆t • a = (w2 - w1)/∆t • a = (1410 - 840 °/sec)/ 1/30 sec • a = (570°/sec)/ 1/30 sec = 17,100 °/sec2 q6 = 150° w1 q7 = 178° w2 q8 = 225°

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