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all about motion that was discuss to our class..

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- 1. Motion<br />Group III<br />GE2-1<br />
- 2. Distance(s)- simply pertains to length (scalar)e.g. 5m, 5km<br />Displacement (s)- change in position specified by magnitude + direction (vector) e.g. 5m, north , 5m, 20˚ S of E.<br />Speed (v) – distance per unit time (scalar)<br />where v= speed<br /> s= distance<br /> t= time<br />
- 3. Velocity ( v )- displacement per unit time<br />where: v = velocity<br /> s = displacement<br /> t = time<br /><ul><li>Acceleration – time rate of change of velocity</li></ul> if <br />Where: a= acceleration<br /> t= time<br /> v0= Initial velocity<br /> vi= final velocity<br />Uniformly Accelerated Motion: the motion in a straight line in which the rate changes uniformly<br />
- 4. Equations <br />1.<br />2.<br />3.<br />4.<br />5.<br />V=average velocity/speed<br />
- 5. Sample problems:<br />1.An airplane taking off from a landing field has a run of 1200ft. If it starts from rest, moves with a constant acceleration and makes he run in 30s, with what velocity in m/s did it take off?<br />
- 6. Illustration:<br />1ft=0.3048m<br />(1)<br />From we can derive <br />
- 7. (2) From we can derive<br />
- 8. 2. A car comes to a stop in 6s from a velocity of 30m/s.<br /> a.) what is its acceleration?<br /> b.) at the same acceleration, how long would it take the car to come to a stop from a velocity of 40 m/s?<br />
- 9. Given: Vo=30m/s<br />t=6s<br />VI=0<br />Sol’n:<br />A.)<br />B.) <br />
- 10. 3. a runner “A” can run the mile race in 4.25min. Another runner “B” requires 4.55mins to finish this distance. If they start out together and maintain their normal speed, how far apart will they be at the end of the race?<br />
- 11. Solution:<br />
- 12.
- 13. Freely falling body- a body which is acted on by no force of appreciable magnitude other than its weight.<br />
- 14. equations<br />1.<br />2.<br />3.<br />4.<br />5.<br />
- 15. Cases <br /> object dropped from a height<br />An object thrown vertically downward<br />S max<br />g=t<br />S max<br />g=t<br />
- 16. An object thrown vertically upward<br />0<br />g=-<br />g =+<br />S max<br />+up<br />+down<br />+up=+down<br />
- 17. Sample Problem<br />A stone is dropped from the edge of a cliff<br /> a.) what is its velocity 3s later<br />b.) how far does it fall in this time?<br />c.) how far will it fall in the next second?<br />
- 18. A.<br /> B.<br />
- 19. C. <br />
- 20. A girl throws the ball 60ft vertically into the air<br />a. how long does she has to wait to catch it on the way down?<br /> b. what is its initial velocity<br /> c. what is its final velocity<br />
- 21. given:<br />solution<br />
- 22. Projectile Motion<br />Projectile motion is an object which is given an initial velocity and after some time allowed to move under the action of gravity<br />t up<br />t down<br />S max<br />g=-<br />g=+<br />(R)<br />
- 23. Trajectory- path followed by projectile<br />Range- horizontal distance<br />t'= total time of flight<br />t'=2 t up -2 t down<br />Note: horizontal comp. of velocity is constant<br />
- 24. Sample problem<br />A canon is elevated at an angle of 45˚. It fires a ball with a speed of 300m/s<br />A.) what height does the ball reach?<br />B.) how long is the ball in the air?<br />C.) what is the range?<br />
- 25. a.) S max (S)<br />b.) t'=?<br />c.) range (R)<br />Solution:<br />b.<br />a.<br />.<br />0<br />
- 26. c.<br />
- 27. A stone is thrown from a window with an initial horizontal velocity of 10m/s. if the window is 20m high, and the ground is level,<br />a. in how many seconds will the stone reach the ground?<br />b. how far in the ground will it reach?<br />
- 28. Required:<br />t=?<br />R=?<br />Solution:<br />a.<br />b.<br />
- 29. Newton’s laws of motion<br />First law: law of Inertia<br />a body at rest remains at rest and a body in motion continues to move in straight line at constant speed, unless an external unbalanced force acts on it.<br />No movement<br />External forces<br />A=B<br />A<br />B<br />Unbalance<br />force<br />rotate<br />A=B<br />
- 30. Second law: Law of Acceleration<br />an external unbalance force acting on an object produces an acceleration in the direction of the net force, an acceleration that is directly proportional to the unbalanced force and inversely proportional to the mass of the body <br />Where:<br />F=force<br />m=mass<br /> a= acceleration<br />Where:<br />w = weight<br />m = mass<br />g = gravity<br />
- 31. Third law: Interaction<br />for every force that a first body exerts upon second body, there is a force equal in magnitude but opposite in direction that a second body exerts upon the first body.<br />
- 32.
- 33. A 100N box is sliding down a frictionless plane inclined at an angle of 30° from the horizontal. Find the acceleration of the box.<br />a=?<br />Note: N is always ┴ to the plane<br />N<br />FBD- free body<br />diagram- shows<br />all the forces acting<br />on the object<br />30°<br />F<br />W<br />
- 34. Solution:<br />
- 35. Uniform circular motion<br /> - motion of an object in a curved or circular motion<br />10 rev. in 5s<br />Frequency, F = the no. of revolution per unit time <br />
- 36. Period, T = the time required for one complete revolution.<br />Where:<br />v= linear velocity<br />r= radius<br />Radian- angle subtended by the arc of a circle whose length is equal to the<br />radius of the same circle.<br />x<br /> r<br />
- 37. Where: ω = angular velocity<br /> = angle turned through<br /> t = time elapse<br />Where: v= linear velocity<br />r= radius<br /> ω=angular velocity<br />Using linear velocity<br />if<br />Centripetal force<br />Using angular velocity<br />
- 38. Sample problem<br />If the radius of the circular path of the stone is 0.5m and its period is 0.5s. What is its constant speed?<br />Given: r= 0.5m<br /> T= 0.5s<br />Required : v=?<br />Solution: <br />
- 39. What is the angular velocity of a stone which makes 10 rev in 5 seconds? The radius of the circular path is 0.5m<br />Given: no. of revolution = 10<br />T= 5s<br />r=0.5m<br />Required: ω=?<br />Solution: <br />
- 40. A mass of 0.5kg is whirled in a horizontal circle of radius 2m. If it makes 5 rev in 5s<br />Find: a. speed<br /> b. acceleration<br /> c. centripetal force<br />Given: m=0.5 kg<br /> r=2m<br /> t=5s<br /> rev=5<br /> T=1s<br />
- 41. Solution:<br />
- 42. End of presentation<br />
- 43. Group III<br />Angelo James Ambion<br />Camille Poniente<br />Junelle Christian Obice<br />Jumel Hernandez<br />John Pierre Pereña<br />Marvin Feranil<br />Marvin Maaño<br />Ryan Sanchez<br />WilfredoPeña<br />

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