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Acceleration

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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.

- 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

- Linear acceleration (a)
- a = ∆ in velocity/ ∆ in time = ∆v/∆t
- UNITSEnglishmetric ft/sec2 m/sec2
- if v = m/sec; t = sec
- m/sec/sec = m/sec2

- 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)

- 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

- 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)

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

- 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/secv2 = 7.9 m/sec
(Q) = 37°∆t = 1.80 sec

- Find: linear acceleration (a)

∆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°

∆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°

- angular acceleration (a)
- a = ∆ in angular velocity/ ∆ in time = ∆w/∆t
- UNITSEnglishmetric rad/sec2 rad/sec2
- deg/sec2 deg/sec2

- 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?

- 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?

- Given: w1 = 260°/secw2 = 1310°/sec
- t1 = 0.20 sec film speed = 30 fr/sec
- 2 frames elapsed

- Find: a
- DIAGRAM:

A

w2

w1

- 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)

- 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!

- 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°

- 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°

- 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°