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Physics II: Lecture 1, Pg 1 AP Physics “Mechanics for Physicists and Engineers” Agenda for Today l 1-D Kinematics (review). çAverage & instantaneous velocity.

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Presentation on theme: "Physics II: Lecture 1, Pg 1 AP Physics “Mechanics for Physicists and Engineers” Agenda for Today l 1-D Kinematics (review). çAverage & instantaneous velocity."— Presentation transcript:

1 Physics II: Lecture 1, Pg 1 AP Physics “Mechanics for Physicists and Engineers” Agenda for Today l 1-D Kinematics (review). çAverage & instantaneous velocity and acceleration çMotion with constant acceleration l Introduction to calculus applications çderivatives and slopes çIntegrals and area

2 Physics II: Lecture 1, Pg 2 Kinematics Problems l 1-D Kinematics çAverage & instantaneous velocity (Chapter2 1,4,5,11- 13,15-17) and acceleration (18,21) (23,24,27,31,35,37,39,40- 1,43) l Motion with constant acceleration(23,24,27,31,35,37,39,40- 1,43) l Free Fall (44,47,49,51,53,56,61,63) l Motion Graphs (66,67,69,70) l Review Phun!!

3 Physics II: Lecture 1, Pg 3 Kinematics l Location and motion of objects is described using Kinematic Variables: l Some examples of kinematic variables. r çposition rvector v çvelocity vvector l Kinematic Variables: çMeasured with respect to a reference frame. (x-y axis) çMeasured using coordinates (having units). Vectors directionmagnitude çMany kinematic variables are Vectors, which means they have a direction as well as a magnitude. V çVectors denoted by boldface V or arrow

4 Physics II: Lecture 1, Pg 4 Motion in 1 dimension r l In general, position at time t 1 is usually denoted r(t 1 ). l In 1-D, we usually write position as x(t 1 ). Since it’s in 1-D, all we need to indicate direction is + or .  Displacement in a time  t = t 2 - t 1 is  x = x(t 2 ) - x(t 1 ) = x 2 - x 1 t x t1t1 t2t2  x  t x1x1 x2x2 some particle’s trajectory in 1-D See text : 2-1

5 Physics II: Lecture 1, Pg 5 1-D kinematics t x t1t1 t2t2  x x1x1 x2x2 trajectory l Velocity v is the “rate of change of position” Average velocity v av in the time  t = t 2 - t 1 is:  t V av = slope of line connecting x 1 and x 2. See text : 2-1

6 Physics II: Lecture 1, Pg 6 l Instantaneous velocity v is defined as: 1-D kinematics... t x t1t1 t2t2  x x1x1 x2x2  t so V(t 2 ) = slope of line tangent to path at t 2. See text : 2-2

7 Physics II: Lecture 1, Pg 7 1-D kinematics... l Acceleration a is the “rate of change of velocity” Average acceleration a av in the time  t = t 2 - t 1 is: l And instantaneous acceleration a is defined as: See text : 2-3

8 Physics II: Lecture 1, Pg 8 Recap l If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time! x a v t t t

9 Physics II: Lecture 1, Pg 9 More 1-D kinematics We saw that v =  x /  t  so therefore  x = v  t ( i.e. 60 mi/hr x 2 hr = 120 mi ) l In “calculus” language we would write dx = v dt, which we can integrate to obtain: l Graphically, this is adding up lots of small rectangles: v(t) t ++...+ = displacement v t 12 60

10 Physics II: Lecture 1, Pg 10 l High-school calculus: l Also recall that l Since a is constant, we can integrate this using the above rule to find: l Similarly, since we can integrate again to get: 1-D Motion with constant acceleration See text : 2-4

11 Physics II: Lecture 1, Pg 11 Recap l So for constant acceleration we find: x a v t t t l From which we can derive: See text : Table 2-1 (p. 33)

12 Physics II: Lecture 1, Pg 12 Problem 1 l A car traveling with an initial velocity v o. At t = 0, the driver puts on the brakes, which slows the car at a rate of a b x = 0, t = 0 abab vovo

13 Physics II: Lecture 1, Pg 13 Problem 1... l A car traveling with an initial velocity v o. At t = 0, the driver puts on the brakes, which slows the car at a rate of a b. At what time t f does the car stop, and how much farther x f does it travel ?? x = x f, t = t f v = 0 x = 0, t = 0 abab vovo

14 Physics II: Lecture 1, Pg 14 Problem 1... l Above, we derived: (a) (b) l Realize that a = -a b l Using (b), realizing that v = 0 at t = t f : find 0 = v 0 - a b t f or t f = v o /a f l Plugging this result into (a) we find the stopping distance:

15 Physics II: Lecture 1, Pg 15 Problem 1... l So we found that l Suppose that v o = 65 mi/hr x.45 m/s / mi/hr = 29 m/s l Suppose also that a b = |g| = 9.8 m/s 2. ç Find that t f = 3 s and x f = 43 m

16 Physics II: Lecture 1, Pg 16 Tips: l Read ! çBefore you start work on a problem, read the problem statement thoroughly. Make sure you understand what information in given, what is asked for, and the meaning of all the terms used in stating the problem. l Watch your units ! çAlways check the units of your answer, and carry the units along with your numbers during the calculation. l Understand the limits ! çMany equations we use are special cases of more general laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).

17 Physics II: Lecture 1, Pg 17 Recap of kinematics lectures l 1-D Kinematics çAverage & instantaneous velocity (Chapter3- 1,3,7,9,11) and and acceleration çMotion Graphs (14,15,17,19) (Ch3 ) l Motion with constant acceleration(Ch3 21,23,27,29,31,35,37 41) l Free Fall (Ch3-) l Free Fall (Ch3-41,43,47,49,51,52) l Review Phun!! () l Review Phun!! (67,69,70 )


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