Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. (Later in the course, we will consider “systems” of particles, and extended bodies) For now, our “particle” or “object” or “body” is represented by a moving single point in space Our questions are, for now, simple: Where is the body? When is it there? Where: position! (in meters, m) When: time! (in seconds, s) Chapter 2: Motion in One Dimension

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Curvilinear motion can take place in 2 or 3 Dimensions, but for now we assume the motion is strictly along a straight line! Where is the body? When is the body there? Where it is: position x, where x is a space displacement from the space origin x = 0 When it’s there: time t, where t is a time interval from the origin of time t = 0 Soon, we’ll ask much more interesting questions about the body’s motion!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What kind of ‘representations’ of motion? Motion diagram: qualitative and pretty, like a movie of the motion Graphical Representation of x as a function of t: x(t): quantitative, visually informative to the trained eye, if labeled carefully Tabular Representation of t and x: include sufficient number of pairs of values to be informative Explicit functional form for x(t): nice for the theorists, and very handy for calculating numbers. Not always obtainable! {show active figure 0201}

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Note: this is a piecewise- Continuous function: OK!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

A tabular representation of this motion Position x [km]Clock timeTime t [min] :00 am0.0 10:10 am10.0 –10.010:20 am20.0 –20.010:30 am30.0 –20.010:40 am40.0 –15.010:50 am50.0 –10.011:00 am60.0

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An explicit function for x(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

A grab-bag of mathematical considerations if two things in an expression are added or subtracted, they are called terms and their dimensions must agree you must make their units agree too when it comes to putting in actual numbers the result will have the same units if two things in an expression are multiplied or divided, they are called factors and their dimensions may differ the result will have units that obey the same algebra of multiplication/division in complicated unit algebra, whatever is on the top of the top, and the bottom of the bottom, is actually on the top whatever is on the bottom of the top, or the bottom of the top, is actually on the bottom

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The velocity concept answer new questions… How fast does object move during a time interval? How fast does the object move at any time t? Crucial notion of the CHANGE in a quantity: The Delta operator  Time interval  t := t 2 – t 1 or  t := t later – t earlier or  t := t final – t initial or  t := t B – t A …. Time interval has the same dimensions as t Often, t initial = 0 s and t final = t (‘present’ time) so if that is the case then  t = t Change in position  x is called displacement

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the average velocity v avg is the slope of the line connecting A and B!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the (instantaneous) velocity Suppose we want to know the velocity v AT A Allow the time interval to shrink by letting t B  t A The displacement also shrinks, as x B := x(t B ) approaches x A :=x(t A )

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Definition of the (instantaneous) velocity In the limit of infinitesimal time interval  t and displacement  x, we have the derivative of x(t): [Show active figure 0203]

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Position–time graph for a particle having an x coordinate that varies with t (quadratically) as x = –4.0 t + 2.0t 2 Note that v(t) is a linear function if x(t) is quadratic!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Properties of the position and the velocity function v(t) is the slope of the tangent line to the graph of x(t) at time t x(t) must be a (piecewise) continuous function (no gaps or holes, passes vertical line test). Why? x(t) may not have any sharp corners, either. If it had them, what would that imply about v(t)? v(t) may be graphed as well. Can it have sharp corners? Must it be continuous? If there is an explicit function x(t), then all of the tricks from calculus may be used to get v(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Speed as distinct from velocity Average speed does not include information about direction of motion Distance is a loose concept: if you walk to the store which is 1km away ‘as the crow flies’, and then return home, your distance may be anywhere from 2 km on up (depending how you wander), but your displacement is ZERO!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Three motion diagrams are shown. Which is a dust particle settling to the floor at constant speed, which is a ball dropped from the roof of a building, and which is a descending rocket slowing to make a soft landing on Mars? A. (a) is ball, (b) is dust, (c) is rocket B. (a) is ball, (b) is rocket, (c) is dust C. (a) is rocket, (b) is dust, (c) is ball D. (a) is rocket, (b) is ball, (c) is dust E. (a) is dust, (b) is ball, (c) is rocket

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem A bicycle moves on a straight path so that its position as a function of time is x(t) = – 200 t+ 50 t 2, where all quantities are in m and s. a)What is the average velocity between t = 2 s and t = 4 s? b) What are the instantaneous velocities at those two times? c) Make graphs of x(t) and v(t)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The next question: how is the velocity changing? Velocity is ‘the rate of change with time’ of position So, acceleration is the rate of change with time of velocity Average acceleration (during a time interval) is defined as The dimensions of acceleration: Length per time, per time a avg is slope of line on a graph of v(t) between two points

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The (instantaneous) acceleration At any instant of time t, the acceleration is the slope of the tangent to the graph of v(t)

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The full set of x(t), v(t) and a(t) graphs, obtained by eyeballing the slopes Note how a(t) may BE discontinuous!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration I there are two ways to work out the set of equations that I call ‘the big five’ for this case. we assume initial time zero and present time t initial position x 0 and present position x(t) := x initial velocity v 0 and present velocity v(t) := v acceleration does not change: call it simply a(t): = A therefore we have (duh!!) a = A (constant) (#1) This came about because of the LINEAR relation between v and t in this special case of constant a

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration II From the definition of the acceleration we have Now insert #3 for v into #2 for x:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Special case: constant acceleration III Finally, it is nice to get an equation that does not involve t at all. To get it, solve #3 for t: Expanding the squared term and combining gives

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Extra-special case: zero acceleration This is synonymous with constant v = v 0 := V the big 5 collapse down into simplicity itself:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A different book’s version of these equations Note that v(t) is a linear function Note that x(t) is a quadratic function Note that v(x) is a square-root function

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: Two railroad engines on parallel tracks are both moving at 25 m/s. Engine A applies its brakes and comes to a stop in 100 seconds (assume constant acceleration), while engine B continues at constant velocity. a) What is the acceleration of both engines? b) How far do both engines travel during that time interval? c) Make three graphs of the three kinematical quantities, using color to distinguish the two engines’ graphs. d) Convert 25 m/s to mi/hr. {show active figures

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: the position function of an object is given by x(t) = – t – 15 ln (2t) where all units are m and s. a)Find an expression for v(t) by taking a derivative. b)Find an expression for a(t) by taking another derivative. c) Graph all three functions, for the time domain 0.5 s< t < 10.0 s.

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Free fall in gravity near the Earth It is known (thanks to Galileo and his contemporaries) That near the Earth, IN THE ABSENCE OF ANY OTHER FORCES BUT GRAVITY, ALL OBJECTS ACCELERATE DOWNWARD AT THE SAME RATE: A = – 9.81 m/s 2 Our convention is to say that A = – g, where g = 9.81 m/s 2 Thus, our 1d coordinate system, called y now, has the positive y direction pointing UP

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens to the big five? We insert a subscript on velocity, to indicate y There is nothing fundamentally different here from motion with uniform acceleration along x!!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. t [s]y [m]  y [m] v y [m/s]  v y [m/s] a y [m/s 2 ] – – 5.0– g – 3.75– 5.0 – g – 6.25– 10.0– 5.0– g – 8.75– 15.0– 5.0– g – 11.25– 20.0– 5.0– g – 13.75– 25.0– 5.0– g X– 30.0– 5.0– g Here,  y now = y later – y now A Table of Free-Fall Kinematics

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example problem: An object is thrown directly upward from the ground at 15 m/s. It climbs vertically, stops for an instant, and returns to the ground. a)How high does it climb? How much time Is required to reach maximum height? b)What are the times for it to be at half of the maximum height? How fast is it moving when it is at that height?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. To find how high it climbs, note that v y = 0 at the top. Use #5 (and for simplicity take g = 10 m/s 2 ): To find the time needed, one could use the above result and plug into #4… but that requires one to solve a quadratic equation for t. Instead, just use #3: #2 also can be used with the above result!! total flight time is 3 seconds!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. To find how the time to be at half of the maximum height, (which is y = m exactly), one can either first find how fast it is moving at that height (using #5), or resort to #4 and bite the quadratic bullet: