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Kinematics: Description of Motion

Kinematics: Description of Motion. Physics September 5, 2008. Kinematics: Description of Motion. The branch of physics concerned with the study of motion and what produces and affects motion is called mechanics. Mechanics is divided into two parts: Kinematics and Dynamics

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Kinematics: Description of Motion

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  1. Kinematics:Description of Motion Physics September 5, 2008

  2. Kinematics: Description of Motion • The branch of physics concerned with the study of motion and what produces and affects motion is called mechanics. • Mechanics is divided into two parts: Kinematics and Dynamics • Kinematics deals with description of motion of objects. • Dynamics analyzes the causes of motion.

  3. Objectives: After completing this lecture, you should be able to: • Define and apply concepts of average and instantaneous velocity and acceleration. • Solve problems involving initial and final velocity, acceleration, displacement, and time. • Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration. • Solve problems involving a free-falling body in a gravitational field.

  4. Uniform Acceleration in One Dimension: • Motion is along a straight line (horizontal, vertical or slanted). • Changes in motion result from a CONSTANT force producing uniform acceleration. • The cause of motion will be discussed later. Here we only treat the changes. • The moving object is treated as though it were a point particle.

  5. B d = 20 m A Distance and Displacement Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below: Distanced is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. (20 m, 40 mi/h, 10 gal)

  6. B x = 12 m, 20o A q Distance and Displacement Displacement is the straight-line separation of two points in a specified direction. A vector quantity: Contains magnitude AND direction, a number,unit & angle. (12 m, 300; 8 km/h, N)

  7. x 8 m,E x = -4 x = +8 12 m,W Distance and Displacement • For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W. Net displacement x is from the origin to the final position: x x = 4 m, W What is the distance traveled? 20 m !!

  8. Examples: The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. The Signs of Displacement • Displacement is positive (+) or negative (-) based on LOCATION. 2 m -1 m -2 m The direction of motion does not matter!

  9. B d t 20 m 4 s s = = A Definition of Speed • Speed is the distance traveled per unit of time (a scalar quantity). d = 20 m s = 5 m/s Not direction dependent! Time t = 4 s

  10. d = 20 m B x=12 m A 20o Time t = 4 s Definition of Velocity • Velocity is the displacement per unit of time. (A vector quantity.) v v = 3 m/s at 200 N of E Direction required!

  11. x2 = 300 m x1 = 200 m Avg. speed 8.33 m/s 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: start Total distance: x = 200 m + 300 m = 500 m Direction does not matter!

  12. t = 60 s xf= -100 m x1= +200 m xo = 0 Average velocity: Example 1 (Cont.)Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters. x0 = 0 m; xf = -100 m Direction of final displacement is to the left as shown. Note: Average velocity is directed to the west.

  13. 14 s A 600 m B 400 m 150 s 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? Total distance/ total time: Average speed is a function only of total distance traveled and the total time required.

  14. Orbit 2 x 104 m/s Light = 3 x 108 m/s Car = 25 m/s Jets = 300 m/s Examples of Speed

  15. Runner = 10 m/s Glacier = 1 x 10-5 m/s Snail = 0.001 m/s Speed Examples (Cont.)

  16. x = 20 m B C A Time t = 4 s Average Speed and Instantaneous Velocity • The averagespeed depends ONLY on the distance traveled and the time required. The instantaneousvelocity is the magn-itude and direction of the speed at a par-ticular instant. (v at point C)

  17. - + + - + The Signs of Velocity • Velocity is positive (+) or negative (-) based on direction of motion. First choose + direction; then vis positive if motion is with that direction, and negative if it is against that direction.

  18. slope x2 Dx Dx Displacement, x x1 Dt Dt t1 t2 Time Average and Instantaneous v Average Velocity: Instantaneous Velocity:

  19. Physics September 9, 2008

  20. Physics QuickySeptember 9, 2008 A jogger jogs from one end to the other of a straight 300-m track in 2.50 min and then jogs back to the starting point in 3.30 min. What was the jogger’s average velocity (in m/s) … • To the far end of the track • Coming back to the starting point • For the total jog

  21. Review “Match Graph” Activity

  22. The direction of accel- eration is same as direction of force. • The acceleration is proportional to the magnitude of the force. Definition of Acceleration • An acceleration is the change in velocity per unit of time. (A vector quantity.) • A changeinvelocity requires the application of a push or pull (force). A formal treatment of force and acceleration will be given later. For now, you should know that:

  23. F a 2F 2a Acceleration and Force Pulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force.

  24. Force t = 3 s + v0= +2 m/s vf= +8 m/s 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.

  25. + F F The Signs of Acceleration • Acceleration is positive (+) or negative (-) based on the direction of force. Choose + direction first. Then accelerationawill have the same sign as that of the force F —regardless of the direction of velocity. a (-) a(+)

  26. slope v2 Dv Dv v1 Dt Dt t1 t2 time Average and Instantaneous a

  27. Force + v1= +8 m/s 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? t = 4 s v2= +20 m/s 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.

  28. Force + t = 4 s v2 = +20 m/s v1= +8 m/s Step 5. Recall definition of average acceleration. Example 3 (Continued):What is average acceleration of car?

  29. + Force E 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.) vf= -5 m/s vo= +20 m/s Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. Step 3. Label given info with + and - signs.

  30. 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, vo=+20 m/s, east (+) Final velocity, vf = -5 m/s, west (-) The change in velocity, Dv = vf - v0 Dv = (-5 m/s) - (+20 m/s) = -25 m/s

  31. vf - vo tf - to Dv Dt aavg= = vo= +20 m/s vf= -5 m/s + Force Dv = (-5 m/s) - (+20 m/s) = -25 m/s E -25 m/s 5 s a = Example 4: (Continued) Acceleration is directed to left, west (same as F). a = - 5 m/s2

  32. C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E 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

  33. x = 0 C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E Signs for Velocity What are the signs (+ or -) of velocity at points B, C, and D? • At B,v is zero - no sign needed. • At C, v is positive on way out and negative on the way back. • At D, v is negative, moving to left.

  34. C D B A + a = - 5 m/s2 vf= -5 m/s vo= +20 m/s Force E Signs for Acceleration What are the signs (+ or -) of acceleration at points B, C, and D? • At B, C, and D, a = -5 m/s, negative at all points. • The force is constant and always directed to left, so acceleration does not change.

  35. Ex 5. Slowing It Down: Average Acceleration A couple in their sport utility vehicle (SUV) is traveling 25 m/s down a straight highway. They see an accident in the distance, so the driver slows down to 10 m/s in 5.0 s. What is the average acceleration of the SUV?

  36. Ex 6. Fast Start, Slow Stop: Motion with Constant Acceleration (At constant acceleration, the previous equation can be written as v = v° + at) A drag racer starting from rest accelerates in a straight line at a constant rate of 5.5 m/s2 for 6.0 s. • What is the racer’s velocity at the end? • If a parachute deployed at this time causes the racer to slow down at a rate of 2.4 m/s2, how long will it take the racer to come to a stop?

  37. Ex 7. Calculating Displacement … how far? Equation 2: x = ½ (v + v°) t Equation 3: x = v°t + ½ a t2 Example: • A car starting from rest accelerates in a straight line at a constant rate of 2.0 m/s2 for 6 seconds. How far does the car travel? • A motorcycle, which set the world record for acceleration, increased speed from rest to 96.0 km/h in 3.07 s. What distance was traveled by the time the final speed was achieved?

  38. AP Physics Homework (9/10) Read Chapter 2 Conceptual Physics (Hewitt) Do Problems 22, 23, 26, 27, 30, 33, 41, 43, 44, 48 Read Chapter 2 in College Physics (Wilson) Do Problems 19, 37, 41, 51, 55, 63, 87, 93, 107, 109

  39. Finding Velocity without TIME Equation 3: v2 = v°2 + 2 a x Ex. A ball is thrown up into the air at an initial velocity of 24 m/s. Assuming it is decelerating at 10 m/s2, how far has it traveled at its maximum height?

  40. g W Earth Acceleration Due to Gravity • Every object on the earth experiences a common force: the force due to gravity. • This force is always directed toward the center of the earth (downward). • The acceleration due to gravity is relatively constant near the Earth’s surface.

  41. Gravitational Acceleration • In a vacuum, all objects fall with same acceleration. • Equations for constant acceleration apply as usual. • Near the Earth’s surface: a = g = 9.80 m/s2 or 32 ft/s2 Directed downward (negative).

  42. Dt y 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 0.495 s. What is the acceleration due to gravity?

  43. Dt y + Acceleration of Gravity: W Experimental Determination of Gravity (y0 = 0; y = -1.20 m) y = -1.20 m; t = 0.495 s Acceleration ais negative because force W is negative.

  44. UP = + Sign Convention:A Ball Thrown Vertically Upward a = - v = 0 y = + a = - v = + y = + y = + a = - • Displacement is positive (+) or negative (-) based on LOCATION. v = - v = - y = 0 a = - y = 0 Release Point • Velocity is positive (+) or negative (-) based on direction of motion. y = -Negative v= -Negative a = - • Acceleration is (+) or (-) based on direction of force (weight). Tippens

  45. Same Problem Solving Strategy Except a = g: • Draw and label sketch of problem. • Indicate + direction and force direction. • List givens and state what is to be found. Given: ____, _____, a = - 9.8 m/s2 Find: ____, _____ • Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

  46. + a = g Example 7:A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s? Step 1. Draw and label a sketch. Step 2. Indicate + direction and force direction. Step 3. Given/find info. a = -9.8 m/s2 t = 2, 4, 7 s vo = + 30m/sy = ?v = ? vo= +30 m/s

  47. Step 4. Select equation that contains y and not v. + a = g 0 vo= 30 m/s Finding Displacement: y = (30 m/s)t + ½(-9.8 m/s2)t2 Substitution of t = 2, 4, and 7 s will give the following values: y = 40.4 m; y = 41.6 m; y = -30.1 m

  48. + a = g vo= 30 m/s Finding Velocity: Step 5. Find v from equation that contains vand not x: Substitute t = 2, 4, and 7 s: v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s

  49. + a = g vo = +96 ft/s Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity vf is zero. To find ymaxwe substitute t = 3.06 s into the general equation for displacement. y = (30 m/s)t + ½(-9.8 m/s2)t2

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