Unit 3 Motion at Constant Acceleration Giancoli Sec

Unit 3 Motion at Constant Acceleration Giancoli, Sec 2 - 5, 6, 8 © 2006, B. J. Lieb

Example 3 -1 Consider the case of a car that accelerates from rest with a constant acceleration of 15 m / s 2 starting at t 1 = 0. We see that velocity increases by 15 m/s every second and is thus a linear function of time. t (s) v ( m/s) a ( m/ s 2 ) 0 0 15 15 2 30 15 3 45 15 4 60 15 5 75 15 Unit 3 - 2

Derivation of Equations for Motion at Constant Acceleration • In the next 4 slides we will combine several known equations under the assumption that the acceleration is constant. • This process is called a derivation. • It will produce four equations connecting displacement, velocity, acceleration and time. • You will use these equations to solve most of the problems in Units 3, 4 and 7. Skip Derivation Unit 3 - 3

Derivations • Much research in Physics requires derivations. • In more advanced courses students are required to be able to perform derivations on tests and homework • In this course you will need to know the initial assumptions, the resultant equations and how to apply them. • You do not need to memorize derivations. • But I could ask you to derive an equation for a specific problem. This is very similar to an ordinary problem without a numeric answer. Unit 3 - 4

Motion at Constant Acceleration - Derivation • Consider the special case acceleration equals a constant: a = constant • Use the subscript “ 0” to refer to the initial conditions • Thus t 0 refers to the initial time and we will set t 0 = 0. • At this time v 0 is the initial velocity and x 0 is the initial displacement. • At a later time t, v is the velocity and x is the displacement • In the equations t 1 t 0 and t 2 t Unit 3 - 5

Motion at Constant Acceleration - Derivation The average velocity during this time is: The acceleration is assumed to be constant From this we can write Unit 3 - 6

Motion at Constant Acceleration - Derivation Because the velocity increases at a uniform rate, the average velocity is the average of the initial and final velocities From the definition of average velocity And thus Unit 3 - 7

Equations for Motion at Constant Acceleration • The book derives one more equation by eliminating time • The notation in the equations has changed • At t = 0, x 0 is the displacement and v 0 is the velocity • At a later time t, x is the displacement and v is the velocity Unit 3 - 8

Example 3 -2. A world-class sprinter can burst out of the blocks to essentially top speed (of about 11. 5 m/s) in the first 15. 0 m of the race. What is the average acceleration of this sprinter, and how long does it take him to reach that speed? (Note: we have to assume a=constant) Unit 3 -9

Graphical Analysis of Linear Motion • Consider the graph of x vs. t. The graph is a straight line, which means the slope is constant. • The slope of the line is the rise (Δy) over the run (Δx). • If we compare this with the definition of velocity, we see that the slope of the x vs. t graph is the velocity. Unit 2 - 10

Graphical Analysis of Linear Motion The graphs describe the motion of a car whose velocity is changing: v is the slope of position vs. time graph. Since the graph is not linear, we draw a tangent line at each point and find slope of the tangent line. Thus a is the slope of velocity vs. time graph. Unit 3 - 11

Example 3 -3: Calculate the acceleration between points A and B and C. Unit 3 - 12

Example 3 -4. A truck going at a constant speed of 25 m/s passes a car at rest. The instant the truck passes the car, the car begins to accelerate at a constant 1. 00 m / s 2. How long does it take for the car to catch up with the truck. When the car catches the truck: How far has the car traveled when it catches the truck? Unit 3 - 13

Example 3 -4: Graphical Interpretation In order to understand the solution to example 4, we can graph the two equations: • • • The graph of the truck (red) is linear because the velocity is constant. The car is accelerating so its graph (blue) is quadratic. The two curves intersect at t = 50 s which agrees with the solution. They intersect at x ~ 1250 m which also agrees. The slopes of the two graphs at t = 50 s indicate that the car is traveling twice as fast as the truck. Unit 3 - 14

Unit 3 Appendix Photo and Clip Art Credits Some figures electronically reproduced by permission of Pearson Education, Inc. , Upper Saddle River, New Jersey Giancoli, PHYSICS, 6/E © 2004. Slide 3 -2 Drawing: Copyright © Pearson Prentice Hall, Inc, reproduced with permission 3 -9 Sprinter: Microsoft Clipart Collection, reproduced under general license for educational purposes. 4 -10 Graphs: Copyright © Pearson Prentice Hall, Inc, reproduced with permission 4 -11 Graph: Copyright © Pearson Prentice Hall, Inc, reproduced with permission 4 -4 Photo: Copyright © Pearson Prentice Hall, Inc, reproduced with permission 4 -4 Graph: Copyright © Pearson Prentice Hall, Inc, reproduced with permission 4 -5, 6, 7 Soccer Kicker: Microsoft Clipart Collection, reproduced under general license for educational purposes. 4 -8, 9 Photo of Leaning Tower of Pisa: Microsoft Clipart Collection, reproduced under general license for educational purposes. Unit 03 - 15

Quadratic Equation A quadratic equation is a polynomial equation of the second degree (one term is squared). The general form of a quadratic equation is where a, b and c are constants and a is not equal to zero. You will need to arrange your equation in this form in order to determine the numeric value of a, b and c. Often this semester, the unknown variable will be t instead of x. A quadratic equation has two solutions or roots: and These solutions are usually combined using the ± symbol: • The solutions can be equal ( if b 2 = 4 a c ). • If they are not equal, you will often need to select the solution that makes sense for the problem. Unit 3 - 16