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Linear Distance and Displacement

Linear Distance and Displacement . Linear Distance The total length of the path traveled by a moving object. Units - Linear displacement and distance are measured in distance units English metric ft m in, miles, etc, cm, mm, µm, km, etc. Linear Distance and Displacement.

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Linear Distance and Displacement

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  1. Linear Distance and Displacement • Linear Distance The total length of the path traveled by a moving object. • Units - Linear displacement and distance are measured in distance units Englishmetric ft m in, miles, etc, cm, mm, µm, km, etc

  2. Linear Distance and Displacement • Linear Distance • The total length of the path traveled by a moving object. • Rectilinear (straight line): • 1 dimension: d = |pos. 2 - pos. 1| • 2 dimensions: Use Pythagorean theorem d = √ (∆x2 + ∆y2) • 3 dimensions: d = √( ∆x2 + ∆y2 + ∆z2)

  3. finish start • Linear Distance – In a 200 m race, the race distance is determined by the actual path covered by the runner.

  4. finish start displacement • Linear Displacement – In a 200 m race, the displacement is the direct distance across the infield from the start to the finish.

  5. A baseball captured on high-speed digital video tape moves from a position of (10,15) to (30,40) as measured in cm. • What is the rectilinear distance traveled? • Given: x1 = 10 cm x2 = 30 cm • y1 = 15 cm y2 = 40 cm • Find: distance (d)

  6. cm Pos 2 (30,40) 40 d • DIAGRAM • Formula: c2 = d2 = a2 + b2 • ∆x = x2 - x1 = 30 - 10 = 20 cm • ∆y = y2 - y1 = 40 - 15 = 25 cm 30 ∆y 20 (10,15) Pos 1 ∆x 10 cm 20 10 40 30 50

  7. Pos 1 (30,40) 40 d • ∆x = 20 cm • ∆y = 25 cm • Formula: d2 = a2 + b2 • d =√ (202 + 252) = √ (400 + 625) = d = √1025 = 32.02 cm 30 ∆y 20 (10,15) Pos 2 ∆x 10 20 10 40 30 50

  8. Angular Distance • A change in angular position --> arc • ∆q = |q2 - q1| q1 = 132° origin ∆q q2 = 260°

  9. Arc (Angular Displacement) • Units: radians, degrees • ∆q = |q2 - q1| • ∆q = 260° - 132° = 128° 132° 0° ∆q 260°

  10. Arc (Angular Displacement) • Joint angle - Allen Johnson

  11. Arc (Angle) • Joint angle - Allen Johnson hip

  12. Arc (angle) • Joint angle Qhip = Qtorso - Qleg Qhip = 26° - (-10°) Qhip = 26° +10° = 36° +26° origin -10°

  13. Angle-> arclength • Susan is performing arm curl exercises holding a 5 kg dumbbell, The distance from her elbow to her palm = 35 cm. If the elbow is flexed through a105° range of motion, find the arclength traced by the dumbbell? • Given: length = 35 cm = r • ∆ q = 105° • Find: arclength (AL) humerus 35 cm elbow 105°

  14. Angle-> arclength • arclength (AL) • AL = r • ∆qrad • AL = 35 cm • 105° • 1 rad/57.3° • AL = 35 cm • 1.83 rad • AL = 64.05 cm 35 cm 105° AL

  15. Linear Velocity • Change in linear distance/time • v = d/∆t • v = (pos. 2 - pos. 1)/∆t note: or d in 2 or 3 D from Pythag. Theorem Units: Englishmetric ft/sec m/sec

  16. Linear Velocity • Average Linear Velocity –total distance divided by time it takes to cover the distance. • EXAMPLE: A runner takes 10 s to run a 100 m race. The average velocity during the race is: v = 100 m/10 s = 10 m/s

  17. Linear Velocity - example • Carl Lewis reached the 40 m mark in the 100 m dash at 4.45 sec. If he reaches the 70 m mark at 7.25 sec., what is his average speed between 40-70 meters? • Given: pos1 = 40 m t1 = 4.45 sec • pos2 = 70 m t2 = 7.25 sec • Find: average velocity (v) 40 m 60 70 m 50

  18. Linear Velocity - example • Formula: v = d/∆t = (pos2 - pos1)/(t2 - t2) • = (70m - 40 m)/(7.25 sec - 4.45 sec) • = 30 m/2.80 sec • = 10.7 m/sec • = 23.9 mph 40 m 60 70 m 50

  19. Problem using velocity equation • A fly ball is hit and caught by Alex Rodriguez in left field. Josh Hamilton of the Texas Rangers is “tagging up” at third base. When Alex Rodriguez catches the ball, he is 280 ft from home plate. Josh Hamilton then takes off for home at an average speed of 25 ft/sec. If Alex Rodriguez throws the ball 0.75 sec later at an average horizontal velocity of 110 ft/sec, does the ball or Josh arrive first? • Given: d L->H = 280 ft vb = 110 ft/sec • tcatch->throw = 0.75 sec • d S->H = 90 ft vS = 25 ft/sec • Find: shortest time (t)

  20. Problem using velocity equation AR • DIAGRAM: • Formula: v = d/∆t • ∆t = d/v • Solution:: AR: tL = d/v + tcatch->throw • tw = 280 ft/110 ft/sec + 0.75 sec • tw = 2.55 + 0.75 sec = 3.30 sec • JH: tS = d/v • tS = 90 ft/25 ft/sec = 3.60 sec • The ball arrived first! JH

  21. Linear Velocity • Instantaneous Linear Velocity – The actual velocity at a given instant during a movement. • EXAMPLE: At the 15 m point in a 100 m race a runner may be moving at 7 m/s while at the 60 m point the velocity may be 11 m/s. This accounts for changes in velocity during the race. Asafa Powell

  22. Angular Velocity • The change in angular position per unit time • w = ∆q/∆t Units: Englishmetric • radians/sec, degrees/sec radians/sec, degrees/sec

  23. Film analysis is performed on a Shane Lechler, former Aggie, now with the Oakland Raiders. At a time 1.25 seconds into the final swing of his right leg, before he makes contact with the football, his upper leg is in an absolute angular position of 110°. His lower leg is a 220°. Two frames later, his upper leg is in an absolute angular position of 127°, and his lower leg is a 262°. If the film speed 30 frames/sec, find the angular velocity of the knee joint. • Given: qUL1 = 110° qLL1 = 220° • qUL3 = 127° qLL3 = 262° t1 = 1.25 sec film speed = 30 fr/sec • Find: w

  24. .Given: qUL1 = 110° qLL1 = 220° •  qUL3 = 127° qLL3 = 262° • t1 = 1.25 sec film speed = 30 fr/sec • Diagram: kn1 = qLL1 - qUL1 = 220 - 110°kn3 = qLL3 - qUL3 = 262 - 127° kn1= 110° kn3= 135° Frame 3 Frame 1 qUL3 = 127° qUL1 = 110° kn3 kn1 qLL1 = 220° qLL3 = 262°

  25. Formula: w = ∆q/∆t • w = (qkn3 - qkn1) / (t2 - t1) ∆t = 1/film speed x ∆frames • ∆t = 1/30 sec/fr • 2 fr = 0.067 sec • t3 = 1.317 sec • Solution: w = (q2 - q1) / (t3 - t1) • w = (135° - 110°) / (0.067 sec) • w = (25°) / (0.067 sec) • w = 373.13°/sec or 6.51 rad/sec

  26. Angle-> arclength • Given: golf club length = 1.1 m • Film speed = 30 fr/sec; consecutive frames • q1 = 196° • q2 = 268° • Find: speed of the • club relative to • the wrist 196° q 268°

  27. Angle-> arclength • solution: AL = r • ∆qrad • v = d/∆t = AL/∆t • ∆q = 268° - 196° • ∆q = 72° • ∆qrad = 72° • 1rad/57.3° • ∆qrad = 1.26 rad • AL = r • ∆qrad • AL = 1.1 m • 1.26 rad • AL = 1.38 m • v = AL/∆t • v = 1.38m/1/30sec = 41.47 m/sec 196° q 268°

  28. Position-time graphs 100 m • position ∆y ∆x 10 sec time Slope = ∆y/∆x = ∆ pos/∆t

  29. Position-time graphs 100 m B • position A ∆y ∆x 10 sec time Slope = ∆y/∆x = ∆ pos/∆t = velocity!!

  30. 100 m dash III 100 m • position II I 10 sec time ∆ Slope (rate of the change in slope) = ∆ velocity/∆time

  31. Locomotion • Locomotion - the ability to move the center of gravity of the body forward using the legs • walking • running

  32. Locomotion Gait cycle - the time and position from when a foot strikes the ground pushes and swings forward until the same foot makes contact with ground again • Stance phase Swing phase Stride length

  33. Locomotion (Gait) Velocity • Locomotion - the ability to move the center of gravity of the body forward using the legs • walking • running • Gait cycle - the time and position from when a foot strikes the ground pushes and swings forward until the same foot makes contact with ground again • stride length: the distance covered by one gait cycle (measure from heel strike to heel strike (units = meters/stride; feet/stride) • stride rate: the frequency of gait cycles per unit time (units = strides/sec) Stride length

  34. Locomotion Velocity • Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • Ex. Stride length = 2 m/stride stride rate = 3 strides/sec • Running velocity = SL • SR = 2 m/stride • 3 strides/sec • v = 6 m/sec Stride length

  35. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride rate ---> power, sprinting • High stride length ----> high efficiency, endurance

  36. Gait Efficiency Assisted in Founding Harvard Fatigue Lab - w/ D.B. Dill Won Nobel Prize for Muscle Metabolism Energy of Shortening = 5 Joules/kg/stride A.V. Hill

  37. Gait Efficiency Energy of Shortening = 5 Joules/kg/stride Mechanical Efficiency = Mechanical Work (J)/Energy (J) V (m/sec) = SL (m/stride) x SR (strides/sec)

  38. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride length ----> high efficiency, endurance • High maximal aerobic capacity (VO2max)

  39. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride rate ---> power, sprinting • High stride length ----> high efficiency, endurance Scott - ultramarathon runner

  40. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride rate ---> power, sprinting • High stride length ----> high efficiency, endurance Badwater Ultramarathon

  41. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High Stride Rate – Power, Sprinting

  42. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride rate ---> power, sprinting • High stride length ----> high efficiency, endurance Justin Gatlin wins the 100m race in Athens, 2004

  43. Running Velocity = Stride length (m/stride) • Stride rate (strides/sec) • High stride rate ---> power, sprinting • High stride length ----> high efficiency, endurance Usain Bolt, Tyson Gay hold 3 fastest times in 100m

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