Chapter 2: Motion a Straight Line

Transcription

1 Formula Memorization: Displacement What is a vector? Average Velocity Average Speed Instanteous Velocity Average Acceleration Instantaneous Acceleration Constant Acceleration Equation (List all five of them) o Derive each one of them Free-fall Acceleration: g= m s 2 = ft s 2 Fundamental Theorem of Calculus o Velocity o Acceleration 1

2 REVIEW: We all know motion intuitively. We move our arms, legs, head, fingers, toes, etc. We walk and run places. We drive places, we fly places. The world is constantly in motion with cars, planes, bikes, satellites, baseballs, footballs, and all sort of activities that we do and participate in. On the atomic level, atoms and molecules move, bounce around and their motions and velocities determine the temperature of things. Asteroids and planets move across the solar systems, stars move along in the galaxies, and galaxies move relative to each other. The study of motion is called kinematics. Kinematics the classification and comparison of motions. 2

3 So far we have briefly touched on displacement as a vector quantity. The official definition is: Displacement x a change in position from position x 1 to position x 2. x = x 2 x 1 Note: is a symbol for Greek uppercase delta. It represents a change in quantity, and it means the final value of that quantity minus the initial value. Another component of motion is velocity. We can describe average velocity as: 3

4 Average velocity (v avg ) a ratio of the displacement x that occurs during a particular time interval t. v avg = x t = x 2 x 1 t2 t 1 Average velocity is a vector quantity, having both magnitude and direction. Details of motion are irrelevant. You can describe average velocity graphically plotting position vs time. In the graph of position x versus time t, v avg is the slope of the line that connects two particular point. A positive average velocity v avg (and slope) tells us that the line slants upwatrds to the right. A negative average velocity v avg (and slope) tells us that the line slants downward to the right. The average velocity v avg always has the same sign as the displacement x because time ( t) is always positive. 4

5 Average Speed (s avg ) involves the total distance covered by a particle and is independent of direction. s avg = total distance total time Speed is always a positive quantity. Direction of motion is irrelevant. EXAMPLE PROBLEM: You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runts out of gasoline and stops. Over the next 30 minute, you walk another 2.0 km farther along the road toa gasoline station. (a) What is your overall displacement from the beginning of your drive to your arrival at the station? Hint: Do not plug straight into the displacement numbers. You have to reason out problems in physics. We will assume you moved in the position direction of the x-axis. What do we know? At first you drove 8.4 km, then you walked 2.0 km. So you traveled 8.4 km km. x = x 2 x 1 We are assuming our origin (home) is 0. So we find that: x = 10.4 km 0 km = 10.4 km. 5

6 (b) What is the time interval t from the beginning of your drive to your arrival at the station? What do we know? We know you walked for 30 minutes. Do we know how long it took you to travel the 8.4 km? No, so lets start there. Note: I will call the time that it took you to drive 8.4 km as t 1, and will call the walking time as t 2. So to find the driving time, we can start with the average velocity formula: v avg = x t = x 2 x 1 t1 Multiply both sides by t 1. Divide both sides by v avg and we end up with the formula: t 1 = x v avg t 1 = 8.4 km 70 km/h = 0.12 h However, don t forget the question. We will label the walk as t 2 = 0.5h Thus the total time was: t total = t 1 + t 2 t total = 0.12 h + 0.5h = 0.62 hour or 37.2 minutes 6

7 (c) What is your average velocity v avg from the beginning of your drive to your arrival at the station? v avg = x 10.4 km = t 0.62 h v avg = 16.8 km h 17 km h Tactic 1: Do you understand the problem? Can you explain the problem? Having the capability to explain a problem is the best test of your ability to understand a problem. To help understand a problem, write down any given data, with units. Identify the unknowns. Attempt to find any connection between the unknown and the given data. Tactic 2: Are the units okay? Be sure to use a consistent set of units, and convert as needed. Make sure the units to your final solution match. Such as the SI unit for velocity is meters per seconds (m/s); therefore, if solving for velocity than the units should also match meters per seconds (m/s). 7

8 Tactic 3: Is your answer reasonable? Does your answer make sense, or is it too large or too small? For instance, if you calculate your cars velocity to be going faster than the speed of light than your answer is far too large. Tactic 4: Reading a Graph Review how to read a graph if you re having any issues. Also, make sure you know the units being used anytime with using a graph. The graph does not make sense if you don t know what units are being used. Graphing is a very important tool in math and physics so it is very beneficial to make sure you understand how to graph well. Pay attention in calculus. (INSTANTANEOUS VELOCITY WILL NOT TESTED ON JUST SHOWING THE CALCULUS PEOPLE WHERE CALCULUS IS COMMON IN PHYSICS) Instantaneous velocity (v ) the velocity at any instant is obtained by shrinking the time interval t closer and closer to the origin point you are looking at (recall your calculus). As t dwindles, the average velocity approaches a limiting value, which is the velocity at that instant. x v = lim t 0 t = dx dt Note: Velocity v at any instant is the slope of the position-time curve at the point representing that instant. Speed is the magnitude of velocity. 8

9 ACCELERATON: Average Acceleration (a ) - when a particle s velocity changes, the particle is said to undergo acceleration. This means that every time you speed up or slow down, you are accelerating. Acceleration has the SI unit of meters per second per second (m/s 2 ) Average Acceleration (a avg ): a avg = v 2 v 1 = v t 2 t 1 t Acceleration is a vector and does has both a magnitude and direction. We feel acceleration all the time. Our bodies can detect acceleration but not necessarily speed. Rollercoaster and other thrill rides are built on the notion of quickly changing the acceleration to generate a thrill. Instantaneous Acceleration (a ) the acceleration at any instant is the rate at which its velocity is changing at that instant. a = dv dt = d2 x dt 2 Graphically, the acceleration at any point is the slope of the curve of the velocity vector v (t) at that point. Acceleration is a vector quantity, having both magnitude and direction. Large accelerations may be expressed in g units, with 1 g = 9.8 m s 2 Tactic 5: An Acceleration s Sign 9

10 If the sign of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases. Constant Acceleration Equations As the name implies, these equations apply and only apply in situation where the acceleration is constant. Important: Please take the time to remember these equations by heart because they will often play a part in further studies. v = v o + at v 2 = v o 2 + 2a (x x o ) x x 0 = v o t at2 x x 0 = vt 1 2 at2 x x o = 1 2 (v o + v)t Note: Derive the Constant Acceleration Equations: You can derive the equation v = v o + at from: a = a avg = v v o t 0 Moreover, you can derive the equation x x 0 = v o t at2 from a combination of: v avg = x x o t 0, v avg = 1 2 (v o + v) and v = v o + at. From equation v = v o + at, x x 0 = v o t at2, and v avg = 1 2 (v o + v), you should be able to derive the reason of the constant acceleration equation. Furthermore, you can derive the equations v = v o + at and x x 0 = v o t at2 by using an indefinite integral. I do recommend trying to derive all of the constant acceleration equations. 10

11 Free-Fall Acceleration: Free-fall acceleration (g) the acceleration of an object towards the Earth. It is independent of the object s characteristics, such as mass, density, or shape; instead it is the same for all objects so that: (Compare two objects falling) g = 9.8 m s 2 = 32 ft s 2 The constant acceleration equations apply to free fall near Earth s surface (either up or down, when the effects of the air can be neglected). For Free-fall: (1) The directions of motion are now along a vertical y-axis instead of the x-axis with the positive direction of y upward. (2) Free-Fall acceleration is negative that is, downward on the y-axis, toward Earth s center and so it has the value g in the equations. Tactic 6: Meaning of Minus Signs We take the acceleration to be negative (-9.8 m s 2 ) in all problems dealing with falling bodies. A negative acceleration means that, as time goes on, the velocity of the body becomes either less positive or more negative. This is true no matter where the body is 11

12 located and no matter how fast or in what direction it is moving. In Sample Problem 2-8, the acceleration of the ball is negative (downward) throughout its flight, whether the ball is rising or falling since the force of gravity is always trying to pull the ball towards the Earth. NOTE: (Please check on this and do not follow this blindly) Going up, g is negative. Coming down, g is positive. Tactic 7: Unexpected Answers Mathematics often generates answers that you may not have thought of as possibilities; DO NOT automatically discard the answers that you get that do not seem to fit. Examine them carefully for physical meaning. 12

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