* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Acceleration
Inertial frame of reference wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Specific impulse wikipedia , lookup
Classical mechanics wikipedia , lookup
Frame of reference wikipedia , lookup
Hunting oscillation wikipedia , lookup
Variable speed of light wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Coriolis force wikipedia , lookup
Faster-than-light wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Fictitious force wikipedia , lookup
Equations of motion wikipedia , lookup
Jerk (physics) wikipedia , lookup
Classical central-force problem wikipedia , lookup
Chapter 2: 1-D Kinematics Paul E. Tippens, Professor of Physics Southern Polytechnic State University Editing by Mr. Gehman © 2007 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 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 body in a gravitational field. 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. Reference Frames and Displacement Any measurement of position, distance, or speed must be made with respect to a reference frame. For example, if you are sitting on a train and someone walks down the aisle, their speed with respect to the train is a few miles per hour, at most. Their speed with respect to the ground is much higher. Reference Frames and Displacement We make a distinction between distance and displacement. Displacement (blue line) is how far the object is from its starting point, regardless of how it got there. Distance traveled (dashed line) is measured along the actual path. 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: Distance is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. (70 m, 40 mi/h, 10 gal) 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. (40 m east; 8 km/h, N) Reference Frames and Displacement d d2 d1 The displacement is written: Left: Right: Displacement is positive. Displacement is negative. 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 D is from the origin to the final position: D = 4 m, W What is the distance traveled? 20 m !! D 8 m,E x = -4 x x = +8 12 m,W 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. 2m -1 m -2 m The direction of motion does not matter! Definition of Speed • Speed is the distance traveled per unit of time (a scalar quantity). A s = 20 m B v= s t = 20 m 4s v = 5 m/s Time t = 4 s Not direction dependent! Examples of Speed Orbit 2 x 104 m/s Light = 3 x 108 m/s Jets = 300 m/s Car = 25 m/s Speed Examples (Cont.) Runner = 10 m/s Glacier = 1 x 10-5 m/s Snail = 0.001 m/s Definition of Velocity • Velocity is the displacement per unit of time. (A vector quantity.) A d = 12 m B D 12 m v t 4s v = 3 m/s, East Time t = 4 s Direction required! Average Speed & Velocity Speed: how far an object travels in a given time interval s d v t t Velocity includes directional information: D d v t t Avg. Speed vs. Avg. Velocity s 100m v t 4s v = 25 m/s Not direction dependent! s 40m v t 4s v = 10 m/s, East Direction required! Speed Example Exercise FloJo, ’88 Olympics d 100.00m m m v 9.48766 9.488 t 10.54s s s Converting to km/hr: m 1km 3600s km v 9.488 34.16 s 1000m 1hr hr 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 s2 = 300 m speed is a function only of total distance start and total time: s1 = 200 m Total distance: s = 200 m + 300 m = 500 m total path 500 m Average speed time 60 s Direction does not matter! Avg. speed 8.33 m/s 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 x0 t = 60 s xf = -100 m x1= +200 m t x0 = 0 m; xf = -100 m 100 m 0 v 1.67 m/s 60 s xo = 0 Direction of final displacement is to the left as shown. Average velocity: v 1.67 m/s, West Note: Average velocity is directed to the west. 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 600 m + 400 m v t A tB 14 s + 150 s 1000 m v 164 s v 6.10 m/s Average speed is a function only of total distance traveled and the total time required. 14 s A 600 m B 400 m 150 s Average Speed and Instantaneous Velocity The average speed depends ONLY on the distance traveled and the time required. A s = 20 m C Time t = 4 s B The instantaneous velocity is how fast and in what direction an object is moving in a particular instant. (v at point C) The Signs of Velocity Velocity is positive (+) or negative (-) based on direction of motion. + - + - + First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction. Velocity in Position-Time Graphs This is a graph of p vs. t for an object moving with const. velocity. If we take the slope of this line (“rise over run”) we get d slope velocity t Notice that the slope of the p-t curve can be (+) or (-). 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). 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. Example of Acceleration + Force t=3s v0 = +2 m/s vf = +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 constant, thus acceleration is constant. Acceleration You are driving in a car. What can you do to cause the car to accelerate? • Speed it up (step on gas) • Slow it down (step on brake) • Change the direction (turn the wheel) The Signs of Acceleration • Acceleration 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. Negative. Acceleration vs. Deceleration There is a difference between negative acceleration and deceleration: Negative acceleration is acceleration in the negative direction as defined by the coordinate system. Deceleration occurs when the acceleration is opposite in direction to the velocity. Acceleration • To calculate acceleration, we use the following formula: aavg v v f v0 t t f t0 • This can also be written as v f v0 at Example Exercises from Notes • Accelerating Plane v km 1 1000m 1hr m a 235 4.35 2 t hr 15.0s 1km 3600s s • An acceleration of +4.35 m/s2 can also be written as +4.35 m/s/s or +4.35 m s-2. This means that every second, the velocity changes by +4.35 m/s Acceleration Which one of these accelerations is the largest? • 60 mi/hr-s • 60 mi/hr-min • 60 mi/hr-hr Ex. Exercises from Notes, cont. • Car Slowing Down v m m 1 m a 33.2 85.5 41.8 2 t s s 1.25s s • Notice that in computing Δv, you always subtract final from initial: v-v0 • Question: An object moving at -25 m/s with an acceleration of +5 m/s/s. What is its velocity after 7 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 4s. What is the average acceleration? + Force t=4s v1 = +8 m/s Step Step Step Step 1. 2. 3. 4. v2 = +20 m/s Draw a rough sketch. Choose a positive direction (right). Label given info with + and - signs. Indicate direction of force F. Example 3 (Continued): What is average acceleration of car? + Force t=4s v1 = +8 m/s v2 = +20 m/s 20 m/s - 8 m/s Step 5. Recall definition a 3 m/s of average acceleration. 4s aavg v v2 v1 t t2 t1 a 3 m/s, rightward 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 E Force 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. 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, v = vf - v0 v = (-5 m/s) - (+20 m/s) = -25 m/s Example 4: (Continued) + vf = -5 m/s aavg = v t E vo = +20 m/s Force v = (-5 m/s) - (+20 m/s) = -25 m/s = vf - vo tf - to a = - 5 m/s2 a= -25 m/s 5s Acceleration is directed to left, west (same as F). Signs for Displacement + D vf = -25 m/s C A vo = +20 m/s E Force B a = - 5 m/s2 Time t = 0 at point A. What are the signs (+ or -) of displacement moving from? A to B B to C C to D + D vf = -25 m/s Signs for Velocity x=0 A vo = +20 m/s C E Force a = - 5 m/s2 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. B Signs for Acceleration + D vf = -25 m/s C A vo = +20 m/s Force B a = - 5 m/s2 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. Distance and Acceleration • A car starting from rest accelerates at 2 m/s2. Which of the following is true? A. The car covers 2 m of distance every second. B. The car covers 2 m more distance than in the previous second (2m, then 4m, then 6m, etc) C. The car covers more and more distance each second (2m, then 4m, then 8m, etc.) Equations for Constant Acceleration To find the distance an object covers while accelerating, use this equation: 1 2 d d0 vot at 2 (2) Notice that, if a =0 and you start from the origin, this equation becomes more recognizable: 0 1 20 d d 0 vot at vot 2 d v t Eqns for Motion at Constant Acceleration We can combine equations (1) & (2) so as to eliminate t: v v 2a ( d d 0 ) 2 2 0 (3) We now have all the equations we need to solve constant-acceleration problems. v vo at 1 2 d d 0 v0t at 2 2 2 v v0 2ad (1) (2) (3) 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? + d vo 5m 8 m/s d = do + vo + vf 2 F vf -2 m/s Careful t=4s t =5m+ 8 m/s + (-2 m/s) 2 (4 s) (Continued) + F d vo vf -2 m/s 5m t=4s 8 m/s d=5m+ d=5m+ 8 m/s + (-2 m/s) 2 8 m/s - 2 m/s 2 (4 s) (4 s) x = 17 m Use of Initial Position d0 in Problems 0 d d0 v0 v f 2 0 t d d 0 v0t at 1 2 0 2 2a(d d0 ) v v v f v0 at 2 f 2 0 If you choose the origin of your x,y axes at the point of the initial position, you can set d0 = 0, simplifying these equations. The do term is very useful for studying problems involving motion of two bodies. Review of Symbols and Units • Displacement (x, xo); meters (m) • Velocity (v, vo); meters per second (m/s) • Acceleration (a); meters per s2 (m/s2) • Time (t); seconds (s) Review sign convention for each symbol Problem Solving Strategy: Draw and label sketch of problem. Set up coordinate system (which way is + and where is the origin) List givens and state what is to be found. Given: ____, _____, _____ (d,v,vo,a,t) Find: ____, _____ Select equation containing one and not the other of the unknown quantities, and solve for the unknown. 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? +400 ft/s v=0 300 ft + F vo X0 = 0 Step 1. Draw and label sketch. Step 2. Indicate + direction and F direction. Example: (Cont.) v=0 +400 ft/s 300 ft + Step 3. List given; find information with signs. List t = ?, even though time was not asked for. F vo X0 = 0 Given: vo = +400 ft/s v=0 x = +300 ft Find: a = ?; t = ? Continued . . . v=0 x +400 ft/s 300 ft + vo F X0 = 0 Step 4. Select equation that contains a and not t. a= -vo2 2d = -(400 ft/s)2 2(300 ft) 0 v2 =vo2 0 + 2a(d -do) Initial position and final velocity are zero. a = - 267 ft/s2 Why isForce the acceleration negative? Because is in a negative direction! 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. g W Earth 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 (usually negative). 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 0.650 s. What is the acceleration due to gravity? t y Experimental Determination of Gravity (y0 = 0; y = -1.20 m) t y = -1.20 m; t = 0.495 s y v0t at ; v0 0 1 2 2 2 y 2(1.20 m) a 2 2 t (0.495 s) Acceleration 2 of Gravity: a 9.79 m/s Acceleration a is negative because force W is negative. y + W Sign Convention: A Ball Thrown Vertically Upward avy== =-0 + avy= ==-++ ya =+-v == • UP = + Release Point Displacement is positive (+) or negative (-) based on LOCATION. vya== =-0- • yv= =a=Negative Negative Tippens Velocity is positive (+) or negative (-) based on direction of motion. • Acceleration is (+) or (-) based on direction of force (Earth pulling on ball). 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. 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. + a=g Step 3. Given/find info. a = -9.8 m/s2 t = 2, 4, 7 s vo = + 30 m/s y = ? v = ? vo = +30 m/s Finding Velocity: Step 4. Find v from equation that contains v and not x: v f v0 at v f 30 m/s (9.8 m/s )t + a=g 2 Substitute t = 2, 4, and 7 s: vo = 30 m/s v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s Finding Displacement: Step 5. Select equation that contains y and not v. 0 y y0 v0t at 1 2 + a=g 2 y = (30 m/s)t + ½(-9.8 m/s2)t2 Substitution of t = 2, 4, and 7 s will give the following values: vo = 30 m/s y = 40.4 m; y = 41.6 m; y = -30.1 m Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity vf is zero. v f 30 m/s (9.8 m/s )t 0 2 30 m/s t ; t 3.06 s 2 9.8 m/s To find ymax we substitute t = 3.06 s into the general equation for displacement. + a=g vo = +96 ft/s y = (30 m/s)t + ½(-9.8 m/s2)t2 Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½(-9.8 m/s2)t2 t = 3.06 s Omitting units, we obtain: + a=g y (30)(3.06) (9.8)(3.06) 1 2 y = 91.8 m - 45.9 m vo =+30 m/s ymax = 45.9 m 2 Summary of Formulas x x0 v0 v f 2 t v f v0 at Derived Formulas: x x0 v0t 12 at 2 2a( x x0 ) v v 2 f 2 0 For Constant Acceleration Only 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. CONCLUSION OF Chapter 6 - Acceleration