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1 – D Motion. We begin our discussion of kinematics (description of motion in mechanics) Simplest case: motion of a particle in 1 – D Concept of a particle : idealization of treating a body as a single point (we get close to doing this by pinpointing the “license plate of a car,” etc.)

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1 – D Motion

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1 d motion l.jpg

1 – D Motion

  • We begin our discussion of kinematics (description of motion in mechanics)

    • Simplest case: motion of a particle in 1 – D

    • Concept of a particle: idealization of treating a body as a single point (we get close to doing this by pinpointing the “license plate of a car,” etc.)

    • 1 – D motion: Only two possible directions

  • We will be working with vector quantities

    • Characterized by both a magnitude AND a direction

    • Scalar quantities have a magnitude but NOT a direction

  • How do we describe motion in physics?

  • Should be able to answer questions, “Where is it?” along with “Where is it going?” and “How fast is it going?”


Position and displacement l.jpg

+x

Position vectors (keep track of position of car)

Displacement vector (change in position from point 1 to point 2)

Position and Displacement

  • Example: Motion of a car along Sandusky St.

(north)

Park Ave. (reference point or origin)

William St.

(point 1)

Central Ave. (point 2)


Slide3 l.jpg

CQ1: Consider the figure below that shows three paths between position 1 and position 2. (Path C is a half circle.) Which path would result in the greatest displacement for a particle moving from position 1 to position 2?

  • Path A

  • Path B

  • Path C

  • All paths would result in the same displacement.


Velocity l.jpg

Stopwatch measures time t1

Stopwatch measures time t2

(average speed = total distance traveled / time to travel that distance)

Velocity

+x

  • Velocity describes speed (magnitude) and direction

    • Need time as well as position information

  • Average velocity:

Park Ave. (reference point or origin)

William St. (point 1)

Central Ave. (point 2)


Velocity5 l.jpg

Stopwatch measures time t2

Stopwatch measures time t1

Velocity

  • Average velocity could be positive or negative

    • Trip from William St. to Central Ave. (previous picture) – assume t1= 0 s and t2= 40 s:

    • Trip from Central Ave. to William St.:

+x

Park Ave. (origin)

William St. (point 2)

Central Ave. (point 1)


Velocity6 l.jpg

Trip #1

Trip #2

vaveis the same for both trips!

t1

t2

Velocity

  • Assume t1 = 0 and t2 = 40 s:

  • Ave. velocity depends only on the total displacement that occurs during time interval Dt= t2 – t1, not on what happens during Dt:

  • x

    Central Ave.

    William St.

    time

    Park Ave.


    Velocity7 l.jpg

    Velocity

    • Note that each line represents how position changes with time and is not representative of a path in space

    • Slope of line between 2 points on this graph gives average velocity between these points

  • Instantaneous velocity = velocity at any specific instant in time or specific point along path

    • Mathematically, it is the limit as Dt 0 of the average velocity:

    • Say we wanted to know velocity at time t1

    • Imagine moving t2 closer and closer to t1

    • Dx & Dt become very small, but ratio not necessarily small


  • Instantaneous velocity l.jpg

    Slope = instant. velocity at timet2

    Slope = instant. velocity at timet1

    t1

    t2

    Instantaneous Velocity

    • Graphically, instantaneous velocity at a point = slope of line tangent to curve at that point

    x

    Central Ave.

    William St.

    time

    Park Ave.


    Velocity9 l.jpg

    Velocity

    • Be careful not to interchange “speed” with “velocity”

      • “speed” refers to a magnitude only

      • “velocity” refers to magnitude and direction

      • For example, I might run completely around a 1–mi. circular track in ¼ hour, with an average speed (round-trip) = dist. traveled / time = 1 mi / 0.25 hr = 4 mph

      • BUT since

      • In general: instantaneous speed = but average speed 

    (I’m right back where I started!)


    Slide10 l.jpg

    CQ2: The graph below represents a particle moving along a line. What is the total distance traveled by the particle from t = 0to t = 10seconds?

    • 0 m

    • 50 m

    • 100 m

    • 200 m


    Acceleration l.jpg

    Slope = average acceleration

    v2

    v1

    t1

    t2

    Acceleration

    • Acceleration describes changes in velocity

      • Rate of change of velocity with time

      • Vector quantity (like velocity and displacement)

      • “Velocity of the velocity”

    • Average acceleration defined as:

      • are the instantaneous velocities at t2 and t1, respectively

    • To find average acceleration graphically:

    v

    t


    Acceleration12 l.jpg

    Slope of tangent line = instantaneous acceleration

    v1

    t1

    Acceleration

    • We can also find the acceleration at a specific instant in time, similar to velocity

    • Instantaneous acceleration:

    • Graphically:

    • BE CAREFUL with algebraic sign of acceleration – negative sign does not always mean that body is slowing down

    v

    t


    Acceleration13 l.jpg

    Acceleration

    • Example: Car has neg. acceleration if slowing down from pos. velocity, or if it’s speeding up in neg. direction (going in reverse)

    • Must compare direction of velocity and acceleration:

    • Be careful not to interchange deceleration with negative acceleration

    Same sign

    Oppositesign

    (for example, going slower in reverse)


    Slide14 l.jpg

    CQ3: Which animation shown in class gave the correct position vs. time graph? (Assume that the positive x direction is to the right.)

    • Animation 1

    • Animation 2

    • Animation 3

    • Animation 4

    PHYSLET #7.1.1, Prentice Hall (2001)


    Constant acceleration l.jpg

    Constant Acceleration

    v

    • For special case of constant acceleration, then v vs. t graph becomes a straight line

      • Note that slope of above line (= acceleration) is the same for any line determined from 2 points on the line or for any tangent line since (for this special case)

      • We can find the instantaneous acceleration from:

     Constant acceleration means constant rate of increase for v

    t


    Constant acceleration16 l.jpg

    Average value of v:

    (since v is a linear function)

    Also, by definition:

    Setting (2) = (3), and using (1), we get:

    Constant Acceleration

    • Now let t1= 0 (can start our clock whenever we want) and t2 =t (some arbitrary time later), and define v(t1= 0)=v0 and v(t2= t) = v

    • Graphical interpretation:

    (takes form of equation of a line:y = b + mx)

    v

    v

    at

    v0

    v0

    t


    Constant acceleration17 l.jpg

    Constant Acceleration

    • We can use equations (1) and (4) together to obtain an equation independent of t and another equation independent of a using substitution, with the result:

    (independent of t)

    (independent of a)


    Constant acceleration18 l.jpg

    Constant Acceleration

    • Graphs of a, v, and x versus time:

    Motion Graphs Interactive


    Slide19 l.jpg

    CQ4: The graph below represents a particle moving along a line. When t = 0, the displacement of the particle is 0. All of the following statements are true about the particle EXCEPT:

    • The particle has a total displacement of 100 m.

    • The particle moves with constant acceleration from 0 to 5 s.

    • The particle moves with constant velocity between 5 and 10s.

    • The particle is moving backwards between 10 and 15 s.


    Interactive example problem seat belts save lives l.jpg

    Interactive Example Problem:Seat Belts Save Lives!

    Animation and solution details given in class.

    ActivPhysics Problem #1.8, Pearson/Addison Wesley (1995–2007)


    Example problem 2 39 l.jpg

    Example Problem #2.39

    • A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s,

    • how fast is the car going at the end of the braking period, and

    • how far has the car gone?

    • Solution (details given in class):

    • 1.5 m/s

    • 32 m


    Example problem 2 43 l.jpg

    Example Problem #2.43

    • A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with the puck. After 3.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2,

    • how long does it take him to catch his opponent, and

    • how far has he traveled in that time? (Assume that the player with the puck remains in motion at constant speed.)

    • Solution (details given in class):

    • 8.2 s

    • 134 m


    Free fall l.jpg

    Free Fall

    • Particular case of motion with constant acceleration: “free-fall” due to gravity

      • Objects falling to the ground

      • Galileo (16th Century) showed that bodies fall with constant downward acceleration, independent of mass (neglecting air resistance and buoyancy effects)

    • Near the Earth’s surface, this acceleration has a constant value ofg 9.8 m/s2

    • We can use all of the constant-acceleration equations for free fall

      • Will use “y” instead of “x” when referring to vertical positions:


    Free fall24 l.jpg

    Free Fall

    +y(up)

    • Sign of depends on choice of +y – axis direction:

      • Choice of axis not important – as long as you are consistent!

      • always points toward the ground, however

     then =

    0

    0

     then =

    +y(down)


    Slide25 l.jpg

    CQ5: A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero?

    • on the way up

    • at the top of the flight path

    • on the way down

    • halfway up and halfway down

    • none of these


    Free fall example problem l.jpg

    +y

    443 m

    • y = y0 + v0t – ½ gt2

    • – 443 m = 0 + 0 – ½ gt2

    • 886 m / g = t2 t2 = 90.4 s2

    • t =  9.5 s t = +9.5 s (neg. value has no physical meaning)

    Free Fall Example Problem

    How long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower in Chicago? (Of course there is a safety net at the bottom.)

    How fast will he be moving just before he hits the net?

    • v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s

      (negative sign means downward direction)


    Slide27 l.jpg

    CQ6: A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 seconds?

    • 9.8 m

    • 19.6 m

    • 39 m

    • 44 m

    • 27 m


    Final comments about 1 d motion l.jpg

    Final Comments About 1–D Motion

    v

    Free fall only

    vtis called the “terminal velocity”

    vt

    • In reality, Tom Petty’s velocity would “tail off” and start reaching a maximum:

    • Other notes about 1–D velocity and acceleration:

      • In general, both velocity and acceleration are not constant in time

      • Can use methods of calculus to generalize equations

      • General formulas reduce to constant acceleration formulas when a(t)= constant =a

      • Remember that formula d = vt only holds when v = constant!

    Including air resistance effects

    t


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