# Motion at Constant Acceleration Giancoli, Sec 2- 5, 6, 8

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Motion at Constant Acceleration Giancoli, Sec 2- 5, 6, 8
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 / s2 starting at t1 = 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/ s2 ) 15 1 2 30 3 45 4 60 5 75 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 t0 refers to the initial time and we will set t0 = 0. At this time v0 is the initial velocity and x0 is the initial displacement. At a later time t, v is the velocity and x is the displacement In the equations t1t0 and t2  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, x0 is the displacement and v0 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 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

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 B and C.
Unit 3- 12

Example 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 / s2. 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

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