Kinematics UCVTS AIT Physics
Kinematics Kinematics the branch of mechanics that deals with the study of the motion of objects without regard to the forces that cause the motion Displacement A vector that points from an object s initial position (Xo) to its final position (X) and has a magnitude equal to the shortest distance between the 2 positions UCVTS AIT Physics
Vector Example A particle travels from A to B along the path shown by the dotted red line This is the distance traveled and is a scalar The displacement is the solid line from A to B The displacement is independent of the path taken between the two points Displacement is a vector UCVTS AIT Physics
Kinematics Kinematics in 1 Dimension We live in a 3-dimensional world, so why bother analyzing 1-dimensional situations? Well, because any translational (straight-line, as opposed to rotational) motion problem can be separated into one or more 1-dimensional problems. Problems are often analyzed this way in physics (and remember throughout your future career; a complex problem can often be reduced to a series of simpler problems). The first step in solving a kinematics problem is to set up a coordinate system. This defines an origin (a starting point) as well as positive and negative directions. We'll also need to distinguish between scalars and vectors (which we have done already last week.remember? I know you do ). A scalar is something that has only a magnitude, like area or temperature, while a vector has both a magnitude and a direction, like displacement or velocity. In analyzing the motion of objects, there are four basic parameters to keep track of. These are time (t) displacement (x or y) velocity (v) acceleration (a). Time is a scalar, while the other three are vectors. In 1 dimension, however, it's difficult to see the difference between a scalar and a vector. The difference will be more obvious in 2 dimensions. UCVTS AIT Physics
Kinematics Kinematics in 1 Dimension Displacement The displacement represents the distance traveled, but it is a vector, so it also gives the direction. If you start in a particular spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have traveled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (xi) : It is a vector that points from an object s initial position to it s final position. Speed and Velocity Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s east towards campus. After exactly one minute you realize that you've left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again. There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number to calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds. Average speed = distance / elapsed time = 360 / 90 = 4 m/s. The average velocity, on the other hand, is given by: x average velocity : v t In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero distance avg speed total time x average velocity : v x is the displacement vector, v has same direction as x t v instantaneous velocity v t is small t v average acceleration : a v is a vector t UCVTS AIT Physics
Kinematics Acceleration An object accelerates whenever its velocity changes. Going back to the example we used above, let's say instead of instantly breaking into a run the moment you turned around: you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period. If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s per second (or, 0.3 m/s2). We can figure out the average velocity during this time. If the acceleration is constant, which it is in this case, then the average velocity is simply the average of the initial and final velocities. The average of 3 m/s west and 6 m/s west is 4.5 m/s west. This average velocity can then be used to calculate the distance you traveled during your acceleration period, which was 10 seconds long. The distance is simply the average velocity multiplied by the time interval, so 45 m. Similar to the way the average velocity is related to the displacement, the average acceleration is related to the change in velocity: the average acceleration is the change in velocity over the time interval (in this case a change in velocity of 3 m/s in a time interval of 10 seconds). The instantaneous acceleration is given by: a v t As with the instantaneous velocity, the time interval is very small (unless the acceleration is constant, and then the time interval can be as big as we feel like making it). On the way out, you traveled at a constant velocity, so your acceleration was zero. On the trip back your instantaneous acceleration was 0.3 m/s2 for the first 10 seconds, and then zero after that as you maintained your top speed. Just as you arrived back at your front door, your instantaneous acceleration would be negative, because your velocity drops from 6 m/s west to zero in a small time interval. If you took 2 seconds to come to a stop, your acceleration is -6 / 2 = -3 m/s2. UCVTS AIT Physics
On October 15, 1997, in the Black Rock Desert, Nevada, the ThrustSSC set a new unlimited world land speed record reaching 763.035mph, or Mach 1.02.
Physics Problem Solving Steps 1. Don t Panic! Every problem has a solution. Well, at least the problems this year in Physics 2. READ the problem, READ the problem! 3. Construct an informative diagram of the physical situation (sketch). 4. Identify and list the given information in variable form (make a table of values) 5. Identify and list the unknown information in variable form. 6. Read the problem again to make sure you are solving for the correct quantity. 7. Identify and list the equation which will be used to determine the unknown information from the known variables. 8. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown or solve for the unknown variable first and then substitute known values (this is generally preferable and easier) 9. Check your answer to ensure that it is reasonable and mathematically correct (include UNITS in you solution and final answer) UCVTS AIT Physics
Kinematics Equations of Kinematics when Acceleration is Constant When the acceleration of an object is constant, calculations of the distance traveled by an object, the velocity it's traveling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined. Note that the equations apply under these conditions: the acceleration is constant the motion is measured from t = 0 the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs. The equations are: v v at f 0 1 x f x0 ( v0 v ) t 2 1 2 x f x0 v0t at 2 v v 2 a( x x ) 2 2 f 0 f 0 UCVTS AIT Physics
Kinematics Applications of the Equations of Kinetics 1. Make a drawing to represent the situation being studied 2. Decide which directions are positive and negative 3. Make a chart and write down all known values and what the question is asking for. 4. Verify that that the given information contains at least 3 of the 5 kinetic variables. 5. If the motion of the object is divided into segments, remember that the final velocity of one segment is the initial velocity of the next segment. Do Example 4 problem in text UCVTS AIT Physics
Kinematics Freely Falling Bodies (Free Fall) Objects falling straight down under the influence of gravity are excellent examples of objects traveling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again. The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has a value of 9.8 m/s2. This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. A typical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this. You throw a ball straight up. It leaves your hand at 12.0 m/s. How high does it go? If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight? How fast is it traveling when you catch it? Origin = height at which it leaves your hand Positive direction = up (a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation: This gives: 0 = 144 + 2 (-9.8) x Solving for x gives x = 7.35 m, so the ball goes 7.35 m high. (b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time. Another way to do it is simply to plug x = 0 into the equation: This gives 0 = 0 + 12 t - 4.9 t2 A factor of t can be canceled out of both terms, leaving: 0 = 12-4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s. (c) The answer for part (c) has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation: v = vo + a t which gives: v = 12-9.8 (2.45) = -12 m/s. UCVTS AIT Physics
Freely Falling Bodies (Free Fall) Do Example 12 (CJ8) Kinematics UCVTS AIT Physics
Kinematics 2D Kinematics Motion can be described in terms of time t and the x and y components of the displacement, acceleration and initial and final velocity vectors. Treat the x and y motion separately (each occurs as if the other was not happening!) Combine x and y motions at the end of the problem using Pythagorean Theorem and trigonometry x component x displacement y a acceleration a x v final velocity v f x v initial velocity v y component 0x 0y t elapsed time t y f y usually just constant velocity : v x t v v a t v v a t f x 0x x f y 0y y x 1 1 ( v0x v f x ) t y ( v0y v f y ) t 2 2 x 1 1 v0 xt a 2 2 xt y v0 yt ayt 2 2 v 2 2 2 2 f x v0x 2axx v f y v0y UCVTS AIT Physics 2ay y
UCVTS AIT Physics
2-Dimensional Motion Definition: motion that occurs with both x and y components. Example: Playing pool. Throwing a ball to another person. Each dimension of the motion can obey different equations of motion.
Solving 2-D Problems Resolve all vectors into components x-component Y-component Work the problem as two one-dimensional problems. Each dimension can obey different equations of motion. Re-combine the results for the two components at the end of the problem.
Sample Problem You run in a straight line at a speed of 5.0 m/s in a direction that is 40 o south of west. a) How far west have you traveled in 2.5 minutes? b) How far south have you traveled in 2.5 minutes?
Sample Problem A roller coaster rolls down a 20 o incline with an acceleration of 5.0 m/s 2 (starting from rest). a) How far horizontally has the coaster traveled in 10 seconds? b) How far vertically has the coaster traveled in 10 seconds?
Projectile Motion Something is fired, thrown, shot, or hurled near the earth s surface. Horizontal velocity is constant. Vertical velocity is accelerated. Air resistance is ignored.
1-Dimensional Projectile Definition: A projectile that moves in a vertical direction only, subject to acceleration by gravity. Examples: Drop something off a cliff. Throw something straight up and catch it. You calculate vertical motion only. The motion has no horizontal component.
2-Dimensional Projectile Definition: A projectile that moves both horizontally and vertically, subject to acceleration by gravity in vertical direction. Examples: Throw a softball to someone else. Fire a cannon horizontally off a cliff. You calculate vertical and horizontal motion.
Horizontal Component of Velocity Is constant Not accelerated Not influence by gravity Follows equation: x = V o,x t
Horizontal Component of Velocity
Horizontal and Vertical
Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.
Sample Problem An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?
Sample Problem Playing shortstop, you throw a ball horizontally to the second baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later. a) How far were you from the second baseman? b) What is the distance of the vertical drop?
Launch angle Definition: The angle at which a projectile is launched. The launch angle determines what the trajectory of the projectile will be. Launch angles can range from -90 o (throwing something straight down) to +90 o (throwing something straight up) and everything in between.
Zero Launch angle v o A zero launch angle implies a perfectly horizontal launch.
General launch angle v o You must begin problems like this by resolving the velocity vector into its components.
A situation to consider A hunter spies a monkey in a tree, takes aim, and fires. At the moment the bullet leaves the gun the monkey lets go of the tree branch and drops straight down. How should the hunter aim to hit the monkey? Aim directly at the monkey Aim high (over the monkey's head) Aim low (below the monkey)
Resolving the velocity Use speed and the launch angle to find horizontal and vertical velocity components V o V o,y = V o sin V o,x = V o cos
Resolving the velocity Then proceed to work problems just like you did with the zero launch angle problems. V o V o,y = V o sin V o,x = V o cos
Sample problem A soccer ball is kicked with a speed of 9.50 m/s at an angle of 25 o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?
Sample problem Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25 o above the horizontal. When the snowballs land, is the speed of A greater than, less than, or the same speed of B? Verify your answer by calculation of the landing speed of both snowballs.
Projectiles launched over level ground These projectiles have highly symmetric characteristics of motion. It is handy to know these characteristics, since a knowledge of the symmetry can help in working problems and predicting the motion. Lets take a look at projectiles launched over level ground.
Acceleration of a projectile y g g g g g x Acceleration points down at 9.8 m/s 2 for the entire trajectory of all projectiles.
Velocity of a projectile y v x v y v x v y v x v y v x v y v x x Notice how the vertical velocity changes while the horizontal velocity remains constant.
Position graphs for 2-D projectiles y y x x t t
Velocity graphs for 2-D projectiles Vy Vx t t
Acceleration graphs for 2-D projectiles ay ax t t
Sample problem A golfer tees off on level ground, giving the ball an initial speed of 42.0 m/s and an initial direction of 35 o above the horizontal. a) How far from the golfer does the ball land?
This is what you build when you have an extra F4 Phantom jet engine and a school bus? A team of hot-rodders in Lincoln, Neb., has gone 320 miles an hour in a jet-powered school bus. According to a CBS News report, the bus uses an engine from an old F4 fighter that puts out about 42,000 horsepower certainly enough to overcome the big yellow vehicle s obvious aerodynamic challenges. For people familiar with extreme motorsports, it is no surprise that people are making high-speed runs in school buses. Indeed, the recent speed run represents an intersection of at least two speed-junkie communities: people who routinely make topspeed runs (usually on salt flats, dry lake beds or long runways) and those who enjoy souping up school buses and other unlikely racing vehicles. There are groups that race school buses and drive them in demolition derbies. They fit right in with organizations for other wild competition involving everything from dragracing boats to lawn tractors. For everyday drivers, just the thought of getting stuck behind a plodding school bus can cause shudders. Of course, the hot jet blast from this bus could ruin your paint and your day.