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Uniformly-Accelerated-Motion- GRADE 9 topic
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- 2. Uniformly Accelerated Motion Uniformly AcceleratedMotion (UAM) is motion of an object where the acceleration is constant. In other words, the acceleration remains uniform; the acceleration is equal to a number and that number does not change as a function of time.
- 4. ACCELERATION What doesit mean? Let’s say the car has an acceleration of .
- 5. Time (s) 01 2 3 4 Velocity (m/s) Acceleration is the rate of change in velocity/speed. 0 4 8 12 16 This is an example of Uniformly Accelerated Motion +4 +4 +4 +4
- 6. Base on thosevalues, the car speed/velocity increases by 4 units in every second. Thus, the acceleration is UNIFORM The car is uniformly accelerating at 4m/s2
- 7. REVIEW. . .REVIEW. . . REVIEW. . .
- 8. How can wedetermine the acceleration of motion? We need to determine the following quantities: 1.Velocity or Speed 2.Time of travel Acceleration (a) a With an S.I Unit of meters-per-second squared ()
- 9. UNIFORMLY ACCELERATED MOTION ISCONSIDERED AN IDEALIZED MOTION. To achieve this, only one and NO other external forces should be affecting the motion
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- 11. FRICTION FORCE CAUSEDBY UNEVEN ROAD DRIVING A CAR FORCE FROM THE CAR ENGINE NON UNIFORMLY ACCELERATED MOTION
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- 13. How can weachieve uniformly accelerated motion? The force causing the motion should be the only force present in that situation.
- 14. FALLING AIR RESITANCE GRAVITATIONAL FORCE EXAMPLE: Inthis situation, we must eliminate the AIR RESISTANCE in order to reach uniformly accelerated motion
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- 17. Describing UAM Where: vf – finalvelocity vi – initial velocity d – Displacement t – time of motion a – Acceleration (Constant) TOOL Kinematics - often referred to as the "geometry of motion“ or “language of motion”.
- 18. How to useKinematics in describing motion with uniform acceleration? TOOL A B C D
- 19. Example 1. A caris moving with an initial velocity of 10 m/s and comes to rest after 8 seconds from initial position. Considering that the car has a constant acceleration of -1.25 m/s2 , how far did the car travelled from its initial to its stopping position? Step 1: Determine the given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the car stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used.
- 20. Step 1: Determinethe given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the ball stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used. Step 3: Substitution and solution.
- 21. Step 1: Determinethe given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the car stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used. Step 3: Substitution and solution.
- 22. Step 1: Determinethe given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the car stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used. Step 3: Substitution and solution. 𝒅=(10 𝑚/ 𝑠) (8 𝑠)+ 1 2 (−1.25 𝑚/ 𝑠 2 )(8 s)2 𝒅=(80 𝑚)+ 1 2 (−1.25 𝑚/ 𝑠 2 )(64 𝑠 2 ) 𝒅=(80 𝑚) −(40 𝑚) 𝒅=𝟒𝟎𝒎
- 23. Step 1: Determinethe given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the car stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used. Step 3: Substitution and solution. (0 𝑚 /𝑠 )2 =(10 𝑚/ 𝑠)2 +2(−1.25 𝑚/ 𝑠 2 ) 𝒅 0=100 𝑚2 /𝑠2 +(− 2.5𝑚 /𝑠2 ) 𝒅 −100 𝑚2 /𝑠2 =(−2.5 𝑚/ 𝑠2 )𝒅 −100𝑚 2 / 𝑠 2 −2.5𝑚/𝑠 2 = (− 2.5𝑚/𝑠 2 )𝒅 − 2.5𝑚/𝑠 2
- 24. Step 1: Determinethe given and the unknown in our situation. vi = 10 m/s; t = 8 s, a= -1.25 m/ Since the car stopped after 8 seconds, then the vf = 0 m/s d = ? Step 2: Identify what TOOL should be used. Step 3: Substitution and solution. (0 𝑚 /𝑠)2 =(10 𝑚/ 𝑠)2 +2(−1.25 𝑚/ 𝑠2 ) 𝒅 0=100 𝑚2 /𝑠2 +(− 2.5𝑚/𝑠2 ) 𝒅 −100 𝑚2 /𝑠2 =(−2.5𝑚/ 𝑠2 )𝒅 −100𝑚 2 / 𝑠 2 −2.5𝑚/𝑠 2 = (− 2.5𝑚/𝑠2 )𝒅 − 2.5𝑚/𝑠 2 100𝑚 2.5 =𝒅 𝟒𝟎𝒎=𝒅
- 26. A ball isbeing thrown vertically upward by a boy.As the ball reaches a maximum height of 20m, the ball momentarily stopped and later returns to the ground. How fast is the ball just after the boy throw it? Consider that there is no air resistance.
- 27. 20m A ball isbeing thrown vertically upward by a boy.As the ball reaches a maximum height of 20m, the ball momentarily stopped and later returns to the ground. How fast is the ball just after the boy throw it? Consider that there is no air resistance.
- 28. A ball isbeing thrown vertically upward by a boy.As the ball reaches a maximum height of 20m, the ball momentarily stopped and later returns to the ground. How fast is the ball just after the boy throw it? Consider that there is no air resistance. What are the given? d = 20m vf = 0m/s a = -9.8m/s2 What is asked? How fast is the ball just after the boy throw it? Vi = ? 20m
- 29. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used?
- 30. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used?
- 31. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used?
- 32. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used?
- 33. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used?
- 34. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? What equation/tool should be used? What is being asked!
- 35. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? SOLUTION PROPER Substitution: (0 𝑚 /𝑠)2 =𝑣𝑖 2 +2 (−9.8 𝑚 /𝑠 2 )(20 𝑚) Simplifying: 0=𝑣𝑖 2 +2(−9.8 𝑚/𝑠2 )(20 𝑚) 0=𝑣𝑖 2 −392 𝑚2 / 𝑠2 − 𝑣𝑖 2 =−392 𝑚2 / 𝑠2 𝑣𝑖 2 =392𝑚2 /𝑠2 √𝑣𝑖 2 =√392𝑚2 /𝑠2 𝒗𝒊=𝟏𝟗.𝟖𝒎/ 𝒔
- 36. d = 20mvf = 0m/s a = -9.8m/s2 Vi = ? 𝒗𝒊=𝟏𝟗.𝟖𝒎/ 𝒔