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REVIEW OF KINEMATICS EQUATIONS FOR HORIZONTAL MOTION

Unit 1. REVIEW OF KINEMATICS EQUATIONS FOR HORIZONTAL MOTION. Text Reference 2.4. Remember: …… To change from km/h to m/s you must divide by 3.6 To change from m/s to km/h you must multiply by 3.6. …..Standard Reference system for direction such as:.

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REVIEW OF KINEMATICS EQUATIONS FOR HORIZONTAL MOTION

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  1. Unit 1 REVIEW OF KINEMATICS EQUATIONS FOR HORIZONTAL MOTION Text Reference 2.4

  2. Remember: …… To change from km/h to m/s you must divide by 3.6 To change from m/s to km/h you must multiply by 3.6 …..Standard Reference system for direction such as: Kinematics Formulae: d= V1t + ½ a t 2 V2=V1 +a xt V22=V12 + 2ad d= V1 + V2 x t 2

  3. Note: The sample problems in this lesson involve objects traveling linearly in the horizontal plane or what is known as in the x direction. In other words, the objects are NOT rising of falling in the Earth’s Gravitational Field. --- x axis or horizontal movement Sample Problem #1: Starting from rest an airplane takes off and accelerates at 175 km/h. From the instant of first movement to take-off a passenger notices that the second hand on her watch sweeps out 32.0 s. What is the minimum length of runway required?

  4. First……175 km/h divided by 3.6 = 48.6 m/s Solution: d= Vi + Vf / 2 x t d= 0.0 m/s + 48.6 m/s / 2 x 32.0 s d= 778 m

  5. Sample Problem #2: Another plane is already traveling at a speed of 11.0 km/h as it enters a runway stretch and requires a distance of 1.22 km to take off if an acceleration of 1.60 m/s 2 is maintained. Determine the time to take-off

  6. Solution:

  7. Sample Problem #3: (OPTIONAL PROBLEM) A car is traveling to the left at 16.0 m/s when suddenly the driver applies the brakes. Every second the brakes are applied, the velocity becomes smaller by 2m/s. • Assuming a smooth application of the brakes, how many seconds • will be required for the car to come to a complete stop? (b) If the car skidded all the way, how long will the skid marks be? (c) If instead of coming to a complete stop, the driver braked for only 2.5 s, what would be the velocity of the car when her foot went back to the accelerator?

  8. Solutions: (a). Notice that the formula used here is the manipulation of: V2= V1 + at

  9. The NEGATIVE sign in front of the 64 tells you that the car skidded 64 meters TO THE LEFT.

  10. Text Practice Problems: Pages 73-74

  11. Activity • Extra Read Text P.56-62 • Text Practice Problems: Pages 73-74

  12. Unit 1 REVIEW OF KINEMATICS EQUATIONS FOR VERTICAL MOTION TEXT REFERENCE 2.4

  13. REMEMBER: ----To change from km/h to m/s, you must DIVIDE BY 3.6 ----To change from m/s to km/h you must MULTIPLY BY 3.6 --- Standard Reference System for direction such as: KINEMATICS FORMULAE: d= V1t + ½ a t2 V2=V1 + at V22=V12 +2ad d= V1 +V2 x t 2

  14. NOTE: The sample problems in this lesson involve objects moving linearly in the VERTICAL plane or in the Y direction. In other words, the objects are rising or falling in the Earth’s Gravitational Field. Y axis or Vertical Movement g= - 9.8 m/s2 Sample Problem: A baseball is “popped” straight upward, leaving a bat at 35 m/s

  15. (a) How high does the ball rise in the air? (b) How much time will the ball be in the air? (c) What is the ball’s position 5 seconds after it leaves the bat? (d) How fast will the ball be moving and in what direction at the 5 second time interval? Given: (a) V1=35 m/s , V2= 0 m/s , g= -9.8 m/s2 , Find: d=?? V22=V12 + 2ad 02=(35m/s)2 + 2(-9.8m/s2)(d) - 1225 m2/s2= -19.6 m/s2(d) -1225 m2/s2 / -19.6 m/s2 = d 62.5 m = d

  16. (b) d= V1 +V2 x t 2 62.5 m= 35 m/s + 0 m/s x t 2 62.5 m= 17.5 m/s x t 62.5 m / 17.5 m/s = t 3.6 s = t However, remember that the time in the air = time up + time down Therefore, Total time in air is 3.6 s(up) + 3.6 s(down) = 7.2 s

  17. (c) Given: V1= 35 m/s, t= 5.0 s, a= -9.8 m/s2 Find Position or d d= V1t +1/2 a t2 = (35m/s)(5.0s) + 1/2(-9.8m/s2)(5.0s)2 = 175 m + -123 m = 52 m

  18. (d) Given: t= 5.0s , d=52 m , a=-9.8 m/s2 , V1= 35 m/s Find: Final velocity or V2 V2=V1 + a t = 35 m/s + -9.8 m/s2 x 5.0 s = 35 m/s + - 49 m/s = - 14 m/s

  19. Activity • Extra Read Text P.62-63 • Handout on “acceleration due to gravity • Text Practice Problems: Pages 73-74

  20. Unit 1 INTRODUCTION TO PROJECTILE MOTION Text Reference 3.2

  21. PROJECTILE MOTION: is the motion of an object fired or thrown at an angle to the horizontal whereby the only force acting on the object is GRAVITY. Also known as Parabolic Trajectory (Curved motion). PROJECTILE: A projectile is an object upon which the only force acting is gravity. TRAJECTORY is the path a moving object follows through space as a function of time. Thus, an object undergoing projectile motion, is moving forward (in the x direction) as it would in outer space and also falling (in the y direction) at the same time due to the Earth’s Gravitational Field.

  22. VECTOR COMPONENTS OF A PROJECTILE A ball’s velocity can be resolved into horizontal and vertical components.

  23. Projectiles follow a PARABOLIC path and travel in an ARC. For the purposes of your studying of projectile motion, the following will be considered: …..air friction will be negligible …in the X direction there is UNIFORM MOTION ( a = 0 m/s 2) … in the Y direction there is UNIFORM ACCELERATION ( a = -9.80 m/s2 )

  24. Projectile Motion Calculations: The most common misconception in projectile motion problems is students forget that they are working with VECTORS Projectile motion is TWO DIMENSIONAL (x and y motion). The key to solving these problems is to break these two dimensional problems into two separate one dimensional parts and then recombine them to produce a final answer. The problem set-up will have a set of givens in the X-direction and another set in the Y-direction.

  25. Timing a Parabolic Trajectory: A and B have the same mass and are at the same height At the same time, A is dropped while B is thrown horizontally Which ball will hit the ground first, A or B?

  26. Watch the following Video

  27. THE ANSWER IS…… No matter how hard you throw B, BOTH WILL STAY IN THE AIR FOR THE SAME AMOUNT OF TIME and thus hit the ground at the same time!!!!!!! TIME is the only things connecting the X and Y components for a projectile

  28. A and B are falling at the same rate ( 9.8 m/s every second ) The initial push only caused B to travel further in the x direction . B still fell the same vertical distance ( d y ) at the same rate ( a y = 9.8 m/s2 ) The VERTICAL and HORIZONTAL components are INDEPENDENT of each other. Thus A and B have the same: vertical distance ( d y ) rate of acceleration ( a y = 9.8 m/s2 ) time to fall (t )

  29. The Equations used in solving projectile motion problems are the same as the equations you worked with in Physics 2204 when you studied Kinematics. d= v1 + v2 x t 2 V 2=V1 + a x t V2x=V1x +ax x t dy= V1y +1/2 ay t2 And so on…… d= V1t + ½ a t2 2ad= V22 – V12 But now we will place x and y notation by the variables according to the dimension we are trying to find information on…either X or Y

  30. Apply the quadratic formula to solve for "Δt":

  31. RANGE RANGE refers to the horizontal displacement Dx = Vxt

  32. Calculations involving projectile motion require some elementary trigonometry: Pythagoras theorem: a2 = b2 + c2 Also, you may at times revisit formulae for kinetic and potential energy: KE=1/2 mv2 PEgrav.=mgh

  33. Solving Problems Involving Projectile Motion • Read the problem carefully, and choose the object(s) you are going to analyze. • Draw a diagram. • Choose an origin and a coordinate system. • Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. • Examine the x and y motions separately.

  34. 3-8 Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

  35. SUMMARY The trajectory of the ball shows the two components of the projectile motion namely V x and V y. V y ( y velocity vector ) gets progressively longer. V x ( x velocity vector ) is the same length throughout the motion ….distance in the X direction is defined as the RANGE Dx = Vxt ….TIME is constant in both the X and Y direction There was no V y initially when ball was kicked from roof….V yi =0

  36. …. a projectile is launched AT AN ANGLE and lands AT the same level as the point of projection

  37. ….the projectile is launched AT AN ANGLE and lands BELOW the point of projection ….for the purposes of this course, you will not be expected to find the launch angle of a projectile

  38. Connections and Careers: Many athletes must have an understanding of projectile motion: Ex: a hockey player may not understand the algebraic nuances of projectile motion, but he must have an innate sense of how projectiles work if he is to shoot a puck into the upper corner of the net. Ex: Any sport where an object is thrown, such as baseball, basketball, and football, involves the RANGE equation (dx=vx x t). Even though players do not think about physics during a game in order to make their plays, after years of practice they have “ calibrated eyeballs” that permit them to know just how to throw a ball.

  39. Unit 1 A golf ball is hit horizontally SOLVING PROJECTILE MOTION EXERCISES: Type 1: HORIZONTAL LAUNCHING Text Reference: 3.3

  40. TYPE 1 -- a projectile that is launched horizontally and lands BELOW the point of projection

  41. 3-7 Projectile Motion It can be understood by analyzing the horizontal and vertical motions separately.

  42. Remember, projectiles that are launched horizontally: the angle θ = 0 degrees with the Horizontal the initial velocity in Y direction is 0.0 m/s (V1y = 0 m/s) the horizontal velocity is constant (Vx is constant) the vertical velocity (Vy) is increasing at a rate of 9.8 m/s2 Time is same for both the x and y dimensions of the motion

  43. Practice exercise 2 As a plane flies horizontally at 65.0 m/s, it releases a package from a height of 1.20 × 103 m. (June 26) (i) What is the horizontal distance the package travels after it is released? ii) What is the final velocity of the package?

  44. SOLUTION

  45. Practice exercise 3 In a laboratory activity, students launch a toy car horizontally off a table with a speed of 3.6 m/s as shown. If a 0.25 m wide target is placed 1.0 m from the base of the table, determine whether the car will hit the target. (August 2007)

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