Chapter 6A. Acceleration. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

Transcription

1 Chapter 6A. Acceleration A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 The Cheetah: : A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds. Photo Vol. 44 Photo Disk/Getty

3 Objectives: After completing this module, 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 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 Distance and Displacement Distance is is the length of of the actual path taken by an object. Consider travel from point A to to point B in in diagram below: A s = 20 m B Distance s 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 Distance and Displacement Displacement is is the straight-line separation of of two points in in a specified direction. D = 12 m, 20 o A B A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 30 0 ; 8 km/h, N)

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

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

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

10 Definition of Velocity Velocity is the displacement per unit of time. (A vector quantity.) s = 20 m A D=12 m 20 o Time t = 4 s B v D t 12 m 4 s v = 3 m/s at 20 0 N of E Direction required!

11 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: s 2 = 300 m start s 1 = 200 m Total distance: s = 200 m m = 500 m Average speed total path 500 m Avg. speed time 60 s 8.33 m/s Direction does not matter!

12 Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time.. In this case, the direction matters. v x f t x x 0 = 0 m; x f = -100 m v m 0 60 s Average velocity: x f = -100 m 1.67 m/s x o = 0 t = 60 s x 1 = +200 m Direction of final displacement is to the left as shown. v 1.67 m/s, West Note: Average velocity is directed to the west.

13 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: xa xb v ta tb 1000 m v 164 s 600 m m 14 s s v 6.10 m/s A B 14 s 625 m Average speed is is a function only of of total distance traveled and the total time required. 142 s 356 m

14 Examples of Speed Orbit Light = 3 x 10 8 m/s 2 x 10 4 m/s Jets = 300 m/s Car = 25 m/s

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

16 Average Speed and Instantaneous Velocity The average speed depends ONLY on the distance traveled and the time required. s = 20 m C A Time t = 4 s B The instantaneous velocity is 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 is positive (+) or negative (-)( based on direction of motion First choose + direction; then v is is positive if if motion is is with that direction, and negative if if it it is is against that direction.

18 Average and Instantaneous v Average Velocity: x x2 x1 vavg t t t 2 1 Instantaneous Velocity: v inst x t ( t 0) x 2 x 1 t x Displacement, x slope t x t 1 t 2 Time

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

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

21 Example of Acceleration + t = 3 s Force v 0 = +2 m/s v f = +8 m/s 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 is constant, thus acceleration is is constant.

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

23 Average and Instantaneous a a avg v v v t t t a inst v t ( t 0) slope v 2 v 1 t v t v t 1 t 2 time

24 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? + v 1 = +8 m/s t = 4 s Force v 2 = +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.

25 Example 3 (Continued): What is average acceleration of car? + v 1 = +8 m/s t = 4 s Force v 2 = +20 m/s Step 5. Recall definition of average acceleration. a avg v v v t t t a 20 m/s - 8 m/s 3 m/s 4 s a 3 m/s, rightward

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

27 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, v o o = +20 m/s, east (+) Final velocity, v f f = -5 5 m/s, west (-)( The change in in velocity, v = v f f - v 00 v = (-5( 5 m/s) - (+20 m/s) = -25 m/s

28 Example 4: (Continued) + Force v f = -5 m/s v o = +20 m/s E v = (-5 m/s) - (+20 m/s) = -25 m/s v v f - v o avg = = t t f - t o a avg a = - 25 m/s 5 s a = - 5 m/s 2 Acceleration is directed to left, west (same as F).

29 Signs for Displacement + C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 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

30 Signs for Velocity + x = 0 C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 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.

31 Signs for Acceleration + C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 What are the signs (+ or -) of acceleration at points B, C, and D? At B, C, and D, D a = -55 m/s, negative at all points. The force is constant and always directed to left, so acceleration does not change.

32 Definitions Average velocity: v avg x x x t t t Average acceleration: a avg v v v t t t

33 Velocity for constant a Average velocity: Average velocity: v avg x xf x t t t f 0 0 v avg v 0 2 v f Setting t o = 0 and combining we have: v v x x 0 f t 0 2

34 Example 5: A ball 5 m from the bottom of an incline is traveling initially at 8 m/s. Four seconds later, it is traveling down the incline at 2 m/s.. How far is it from the bottom at that instant? + x F 5 m v o v f -2 m/s 8 m/s t = 4 s Careful v o + v f x = x o + t 2 8 m/s + (-2 m/s) = 5 m + (4 s) 2

35 (Continued) + x F 5 m v o 8 m/s v f -2 m/s t = 4 s 8 m/s + (-2 m/s) x = 5 m + (4 s) 2 8 m/s - 2 m/s x = 5 m + (4 s) 2 x = 17 m

36 Constant Acceleration Acceleration: a avg v vf v t t t f 0 0 Setting t o = 0 and solving for v, we have: v v at f 0 Final velocity = initial velocity + change in velocity

37 Acceleration in our Example v v at a f vf 0 v t 0 5 m + F x v o 8 m/s v -2 m/s t = 4 s a ( 2 m/s) ( 8 m/s) a = m/s 2 4 s 2 m/s What The force is the changing meaning of speed negative is down sign plane! for a? 2

38 Formulas based on definitions: v v x x 0 f t 0 f 0 2 Derived formulas: v v at x x v t at f 2 x x v t at ax ( x) v v f For constant acceleration only

39 Use of Initial Position x 0 in Problems. 0 v v x x 0 f t x x v t at 1 0 f 2 x x v t at ax ( x) v v f v v at f If If you choose the origin of of your x,y axes at at the point of of the initial position, you can set x 00 = 0, 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies.

40 Review of Symbols and Units Displacement(x, x o o ); ); meters (m)( Velocity(v, v o o ); ); meters per second ((m/s) Acceleration (a); meters per s 2 2 (m/s 2 ) Time(t); seconds (s)( Review sign convention for each symbol

41 The Signs of Displacement Displacement is is positive (+) or negative (-)( ) based on LOCATION. 2 m -11 m -22 m The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION.

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

43 Acceleration Produced by Force Acceleration is is (+)( ) or (-)( ) based on direction of force (NOT based on v). F F a(-) a(+) A push or pull (force) is necessary to change velocity, thus the sign of a is same as sign of F. More will be said later on the relationship between F and a.

44 Problem Solving Strategy: Draw and label sketch of problem. Indicate + direction and force direction. List givens and state what is to be found. Given:,, (x,v,v o,a,t) Find:, Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

45 Example 6: A airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration? v = ft + F +400 ft/s v o X 0 = 0 Step 1. Draw and label sketch. Step 2. Indicate + direction and F direction.

46 Example: (Cont.) v = ft + F +400 ft/s v o X 0 = 0 Step 3. List given; find information with signs. List t =?, even though time was not asked for. Given: v o = +400 ft/s v = 0 x = +300 ft Find: a =?; t =?

47 Continued... x v = ft + F Step 4. Select equation that contains a and not t. -v 2 o a = = 2x +400 ft/s v o X 0 = a(x -x o ) = v 2 -v o 2 -(400 ft/s) 2 Initial position and final velocity are a 2(300 ft) = ft/s 2 zero. Because Why is Force the acceleration is in a negative negative? direction!

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

49 Gravitational Acceleration In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual. Near the Earth s s surface: a = g = 9.80 m/s 2 or 32 ft/s 2 Directed downward (usually negative).

50 Experimental Determination of Gravitational Acceleration. t The apparatus consists of a device which measures the time required for a ball to fall a given distance. y Suppose the height is 1.20 m and the drop time is recorded as s. What is the acceleration due to gravity?

51 Experimental Determination of Gravity (y 0 = 0; y = m) y = m; t = s 1 2 y v0t 2 at ; v0 0 a Acceleration of Gravity: 2y 2( 1.20 m) 2 2 t (0.495 s) 2 a 9.79 m/s Acceleration a is negative because force W is negative. y W t +

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

53 Same Problem Solving Strategy Except a = g: g Draw and label sketch of problem. Indicate + direction and force direction. List givens and state what is to be found. Given:,, a = m/s 2 Find:, Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

54 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, s 4 s, and 7 s? s Step 1. Draw and label a sketch. Step 2. Indicate + direction and force direction. Step 3. Given/find info. a = g + a = -9.8 ft/s 2 t = 2, 4, 7 s v o = + 30 m/s y =? v =? v o = +30 m/s

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

56 Finding Velocity: Step 5. Find v from equation that contains v and not x: v v at f 0 a = g + v f 30 m/s ( 9.8 m/s ) t Substitute t = 2, 4, and 7 s: 2 v o = 30 m/s v = m/s; v = m/s; v = m/s

57 Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity v f is zero. v f 2 30 m/s ( 9.8 m/s ) t 0 a = g + t 30 m/s ; t 3.06 s m/s To find y max we substitute t = 3.06 s into the general equation for displacement. v o = +96 ft/s y = (30 m/s)t + ½(-9.8 m/s 2 )t 2

58 Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½(-9.8 m/s 2 )t 2 t = 3.06 s a = g + Omitting units, we obtain: y (30)(3.06) ( 9.8)(3.06) 1 2 y = 91.8 m m 2 v o =+30 m/s y max = 45.9 m

59 Summary of Formulas v v x x 0 f t 0 v v at f 0 2 Derived Formulas: x x v t at f 2 x x v t at ax ( x) v v f For Constant Acceleration Only

60 Summary: Procedure Draw and label sketch of problem. Indicate + direction and force direction. List givens and state what is to be found. Given:,, Find:, Select equation containing one and not the other of the unknown quantities, and solve for the unknown.

61 CONCLUSION OF Chapter 6 - Acceleration