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4. 4 Fundamental Quantities and Units Fundamental QuantityMKS Unit * Length(Meters) * Mass(Kilograms) * Time(Seconds)

5. 5 Motion The symbol “Δ” means “Change in” QuantitySymbolDefinitionUnits Distancedm Speedvd/Δtm/s AccelerationaΔv/Δtm/s 2

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7. 7 The speed changes by 4 m/sec every second

8. 8 After 1 sec v = 7 m/sec

9. 9 After 1 sec v = 7 m/sec After 2 sec v = 11 m/sec

10. 10 After 1 sec v = 7 m/sec After 2 sec v = 11 m/sec After 3 sec v = 15 m/sec

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15. 15 Motion An object is thrown vertically upward from the surface of the earth in the absence of air resistance. Note: this can be accomplished, for example, in an experimental facility from which the atmosphere has been removed. 1. What is the acceleration of the object on its way up? 2. What is the acceleration of the object on its way down? 3. If the object is thrown vertically, it will stop at the highest point in its motion. What is the acceleration of the object at the moment that it stops?

16. 16 Motion Air Hockey There is no friction between the puck and the table surface

17. 17 Motion Free Fall Tower Drop distance - dd yHeight above base - y 100 m Assumption: No air resistance

18. 18 Motion 1. Assume the air hockey puck moves with an initial speed of 10 m/s. What is its position (assume it starts a 0 m) after 1 s, 2 s, 3 s, 4 s and 5 s? 2. Assume the Free Fall Tower is 100 m high. What is the drop distance and the height above the base after 1 s, 2 s, 3 s, 4 s and 5 s? Assume g = 10 m/s 2.

19. 19 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed?

20. 20 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed?

21. 21 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? d = ½ a t 2

22. 22 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? d = ½ a t 2 d = v t

23. 23 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Hmmmmmmmmmmmmmm, we have relationships between distance and time, and speed and time, but no direct relationship between speed and distance.

24. 24 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Hmmmmmmmmmmmmmm, we have relationships between distance and time, and speed and time, but no direct relationship between speed and distance. Solution: d = ½ a t 2

25. 25 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Hmmmmmmmmmmmmmm, we have relationships between distance and time, and speed and time, but no direct relationship between speed and distance. Solution: d = ½ a t 2 and therefore t = SQRT ( 2 d /a) = SQRT (2*10 m / 4.95 m/s 2 ) = 2.01 s

26. 26 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Solution: Then, v = a t

27. 27 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Solution: Then, v = a t = ( 4.95 m/s 2 ) * (2.01 s) = 10 m/s = 22 mph

28. 28 Motion Example: Carl Lewis won the 100 m sprint at the Barcelona Olympics. Data collected during his run shows that he accelerated at a constant acceleration of 4.95 m/s 2 for the first 10 meters of the run. At this point, he reached his maximum speed, which remained fairly constant throughout the rest of the sprint. What was his maximum speed? Solution: Then, v = a t = ( 4.95 m/s 2 ) * (2.02 s) = 10 m/s = 22 mph Moral of the story: The equations of motion can be combined, if done carefully

29. 29 Motion Special note for fancy people:

30. 30 Motion Special note for fancy people: v = a t Or t = v / a

31. 31 Motion Special note for fancy people: v = a t Or t = v / a Then d = ½ a t 2 = ½ a (v / a) 2 = v 2 / 2 a

32. 32 Motion Special note for fancy people: v = a t Or t = v / a Then d = ½ a t 2 = ½ a (v / a) 2 = v 2 / 2 a So v 2 = 2 a d

33. 33 Motion Special note for fancy people: v = a t Or t = v / a Then d = ½ a t 2 = ½ a (v / a) 2 = v 2 / 2 a So v 2 = 2 a d This is an “additional” law relating speed, acceleration and distance.

34. 34 Motion Special note for fancy people: v = a t Or t = v / a Then d = ½ a t 2 = ½ a (v / a) 2 = v 2 / 2 a So v 2 = 2 a d This is an “additional” law relating speed, acceleration and distance. Later, we will learn this is a statement of “Conservation of Energy”