2. The Cheetah : A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds.
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11. Example 1. A runner runs 200 m, east, then changes direction and runs 300 m, west . If the entire trip takes 60 s , what is the average speed and what is the average velocity? Recall that average speed is a function only of total distance and total time : Total distance: s = 200 m + 300 m = 500 m Direction does not matter! start Avg. speed= 8 m/s s 1 = 200 m s 2 = 300 m
12. Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time . In this case, the direction matters. x o = 0 m; x = -100 m Direction of final displacement is to the left as shown. Note: Average velocity is directed to the west. x o = 0 t = 60 s x 1 = +200 m x = -100 m Average velocity:
13. Example 2. A sky diver jumps and falls for 600 m in 14 s. After chute opens, he falls another 400 m in 150 s. What is average speed for entire fall? Average speed is a function only of total distance traveled and the total time required. Total distance/ total time: 625 m 356 m 14 s 142 s A B
14. Examples of Speed Light = 3 x 10 8 m/s Orbit 2 x 10 4 m/s Jets = 300 m/s Car = 25 m/s
18. Acceleration and Force Pulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force. F a 2F 2a
19. Example of Acceleration The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s . Each second the speed changes by 2 m/s . Wind force is constant, thus acceleration is constant. + v = +8 m/s v 0 = +2 m/s t = 3 s Force
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21. Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s . What is average acceleration? Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right). Step 3. Label given info with + and - signs. Step 4. Indicate direction of force F. t = 4 s v = +20 m/s + v o = +8 m/s Force
22. Example 3 (Continued): What is average acceleration of car? Step 5. Recall definition of average acceleration. + v o = +8 m/s t = 4 s v = +20 m/s Force
23. Example 4: A wagon moving east at 20 m/s encounters a very strong head-wind, causing it to change directions. After 5 s , it is traveling west at 5 m/s . What is the average acceleration? (Be careful of signs.) Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. v o = +20 m/s v = -5 m/s Step 3. Label given info with + and - signs. + Force E
24. Example 4 (Cont.): Wagon moving east at 20 m/s encounters a head-wind, causing it to change directions. Five seconds later, it is traveling west at 5 m/s . What is the average acceleration? Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity, v = v f - v 0 v = (-5 m/s) - (+20 m/s) = -25 m/s
25. Example 4: (Continued) = - 5 m/s 2 Acceleration is directed to left, west (same as F). = -25 m/s 5 s + Force v o = +20 m/s v = -5 m/s E v = (-5 m/s) - (+20 m/s) = -25 m/s
26. Signs for Displacement Time t = 0 at point A . What are the signs (+ or -) of displacement at B , C , and D ? At B, x is positive , right of origin At C , x is positive , right of origin At D , x is negative , left of origin + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D
33. Graphical Analysis Average Velocity: Instantaneous Velocity: x t x 2 x 1 t 2 t 1 x t Time slope Displacement, x
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35. Average velocity for constant a : setting t o = 0 combining both equations: For constant acceleration:
36. Formulas based on definitions : Derived formulas : For constant acceleration only
37. Example 6: An airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration? Step 1. Draw and label sketch. Step 2. Indicate + direction Δ x = 300 ft v o = 400 ft/s v = 0 +
38. Example: (Cont.) Step 3. List given; find information with signs. Given: v o = 400 ft/s - initial velocity of airplane v = 0 - final velocity after traveling Δ x = +300 ft Find: a = ? - acceleration of airplane Δ x = 300 ft v o = 400 ft/s v = 0 +
39. Step 4. Select equation that contains a and not t . v 2 - v o 2 = 2 a Δ x a = - 300 ft/s 2 Why is the acceleration negative? Because Force is in a negative direction which means that the airplane slows down Given: v o = +400 ft/s v = 0 Δ x = +300 ft 0 a = = - v o 2 2x -(400 ft/s) 2 2(300 ft)
40. Example 5: A ball 5.0 m from the bottom of an incline is traveling initially at 8.0 m/s . Four seconds (4.0 s) later, it is traveling down the incline at 2.0 m/s . How far is it from the bottom at that instant? Given: d = 5.0 m - distance from initial position of the ball v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find: x = ? - distance from the bottom of the incline 5.0 m Δ x 8.0 m/s -2.0 m/s t = 4.0 s +
41. x = 17.0 m Given: d = 5.0 m v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find: x = ? - distance from the bottom of the incline Solution: where
42. Acceleration in our Example a = -2.50 m/s 2 What is the meaning of negative sign for a ? The force changing speed is down plane! 5 m x 8 m/s -2 m/s t = 4 s v o v + F
43. Use of Initial Position x 0 in Problems. If you choose the origin of your x,y axes at the point of the initial position, you can set x 0 = 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies. 0 0 0 0
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51. Experimental Determination of Gravitational Acceleration. The apparatus consists of a device which measures the time required for a ball to fall a given distance. Suppose the height is 1.20 m and the drop time is recorded as 0.650 s. What is the acceleration due to gravity? y t
52. Experimental Determination of Gravity (y 0 = 0; y = -1.20 m) y = -1.20 m; t = 0.495 s Acceleration a is negative because force W is negative. y t Acceleration of Gravity: + W
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55. Example 7: A ball is thrown vertically upward with an initial velocity of 30.0 m/s . What are its position and velocity after 2.00 s , 4.00 s , and 7.00 s ? Find also the maximum height attained v o = +30.0 m/s Given: a = - Δ 9.8 m/s 2 v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Find: Δ y = ? – displacement v = ? - final velocity After those three “times” Δ y = ? – maximum height a = g +
56. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:
57. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:
58. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For maximum height, v = 0 (the ball stops at maximum height):
Displacement is based on position relative to the origin Speed has no positive or negative Velocity is based on the direction of the motion Acceleration is based on the direction of the force
1bsn1 jan 11
V f = V i + at X f = X o + V o + at 2 V f 2 = V o 2 + 2aX f 1BSN6 JAN26
a = -266.6666.. ≈ -267 ft/s 2
X = 17 m
A = -2.50 m/s2
V f = V i + at X f = X o + V o t+ at 2 V f 2 = V o 2 + 2aX f
