Motion in One Dimension

1
CHAPTER 2

• Motion in One Dimension

2
Kinematics

• The branch of physics concerned with the study of
  the motion of an object and the relationship of
  this motion to such physical concepts as force
  and mass is called dynamics.
• The part of dynamics that describes motion
  without regard to its cause is called kinematics.
• In order to describe the motion of an object, you
  must be able to specify its position at all times
  using some convenient coordinate system and a
  specified origin.
• A frame of reference is a choice of coordinate
  axes that defines your starting point for
  measuring any quantity.

3
Displacement

• Example
• Consider a jogger moving along the x axis, where
  the initial position of the jogger is xi and the
  final position is xf.
• The displacement of the jogger ?x is defined as
  his change in position.
• As he moves from one position to another, his
  displacement is given by the difference between
  his final and initial positions, or xf xi.
• ?x xf - xi

4
Displacement

• From this definition, we see that ?x is positive
  is xf is greater than xi and negative if xf is
  less than xi.
• If the jogger moves from initial position of xi
  3 m to a final position of xf 15 m, his
  displacement is
• ?x 15 m 3 m
• ?x 12 m

5
Displacement

• Displacement is an example of a vector quantity.
• In general, a vector quantity is a physical
  quantity that must be characterized by both a
  magnitude and a direction.
• By contrast, a scalar quantity is one that has a
  magnitude but not direction.
• Ex Mass and temperature.

6
Displacement

• Example
• If a truck moves from an initial position of 10 m
  to a final position of 80 m, what is its
  displacement?

• ?x 80 m (10 m) 70 m
• In this case, the displacement has a magnitude of
  70 m and is directed in the positive x direction,
  as indicated by the plus sign of the result.
• Sometimes the sign is omitted, but a result of
  70 m for the displacement is understood to be the
  same as 70 m.

7
Displacement

• Example
• If the truck moves to the left from an initial
  position of 80 m to a final position of 20 m,
  what is its displacement?

• ?x 20m (80 m) -60 m
• The minus signs in this result indicates that the
  displacement is a vector quantity pointing in the
  negative x direction.

8
Displacement

• Any vector quantity is usually represented by an
  arrow.
• The length of the arrow represents the magnitude
  of the displacement, and the head of the arrow
  indicates its direction.
• We will learn more about diagramming vectors as
  arrows in Chapter 3.
• Do not confuse displacement with distance.
• The displacement of an object is not the same as
  the distance it travels.
• For example, a model rocket fired straight upward
  and returning to the firing point travels some
  distance, but its displacement is zero.

9
Average Velocity

• The motion of an object is completely known it
  its position is known at all times.
• Consider again the truck from the previous
  examples, moving along a highway (the x axis).
• Let the trucks position be xi at some time ti,
  and let its position be xf at some time tf.
• In the time interval ?t tf ti, the
  displacement of the truck is ?x xf xi.
• The average velocity during some time interval is
  defined as the displacement ?x divided by the
  time interval which the displacement occurred.
• v ?x/?t (xf xi)/(tf ti)

10
Average Velocity

• The average velocity of an object can be either
  positive or negative, depending on the sign of
  the displacement.
• The time interval ?t is always positive.
• For the previous examples we did concerning the
  truck, the average velocity for the first one
  would be positive, and the average velocity for
  the second one would be negative.

11
Average Velocity

• Example
• Suppose a friend tells you that she will be
  taking a trip in her car and will travel at a
  constant rate of 55 mph directly northward for 1
  hour. Where will she be at the end of the trip?
• v ?x/?t
• 55 mph ?x/1 hour
• ?x ? 60 miles north

12
Average Velocity

• Example
• Suppose a truck moves 70 m to the right is 5.0
  seconds. What is the average velocity of the
  truck?
• v 70 m/5.0 s
• v 14 m/s
• Note that the units of average velocity are units
  of length divided by units of time.
• Theses are meters per second (m/s) in SI.
• Other units for velocity might be feet per second
  (ft/s) in the U.S. customary system or
  centimeters per second (cm/s) in the metric
  system.

13
Average Velocity

• Example
• Let us assume that we are watching a drag race
  from the Goodyear blimp.
• In one run we see a car follow the straight-line
  path from point P to point Q, during the time
  interval ?t, and in a second run a car follows
  the curved path during the same time interval.

• If you examine the definition of average velocity
  carefully, you will see that the two cars had the
  same average velocity.
• This is because they had the same displacement
  (xf xi) during the same time interval ?t.

14
Average Velocity

• Example
• The figure below shows the unusual path of a
  confused player.
• He receives a kickoff at his own goal, runs
  downfield to within inches of a touchdown, and
  then reverses direction to race backward until he
  is tackled at the exact location where he first
  caught the ball.
• During this run, what is the total distance he
  travels, his displacement, and his average
  velocity in the x direction?

15
Graphical Interpretation of Velocity

• If a car moves along the axis from point A to B
  to C, and so forth, we can plot the position of
  these points as a function of the time elapsed
  since the start of motion.
• This gives a position-time graph such as the one
  below.

16
Graphical Interpretation of Velocity

• Note that the motion is along a straight line,
  yet the position-time graph is not a straight
  line.
• The straight line connecting any two-points on
  this graph, such as point A and D, provides us
  with a geometric interpretation of average
  velocity.
• The slope of the line is ?x divided by the time
  interval for the motion, ?t.
• Slope (Change in y)/(Change in x)
• Therefore, the average velocity of an object
  during the time interval ?t is equal to the slope
  of the straight line joining the initial and
  final points on a graph of the position of the
  object versus time.

17
Graphical Interpretation of Velocity

• Example
• Suppose the position of the car is recorded for
  every second.
• From the data points on the graph below, we see
  that the car first moves in the positive x
  direction as it travels from point A to B to C,
  reaches a maximum position of about 95 m from the
  origin of motion at time t 2.5 s, and then
  reverses its direction as it heads back toward
  the origin of motion.

18
Graphical Interpretation of Velocity

• In the first 3.0 s of its motion, as it travels
  from point A to D, its displacement
• ?x 90 m (50 m) 40 m.
• Thus, its average velocity in this interval,
  which equals the slope of the blue line is
• v ?x/?t (40 m)/(3.0 s) 13 m/s
• If we used this procedure to find the average
  velocity between two other points, we would most
  likely get a different value.
• For example, the average velocity between points
  C and F is negative and has a value of about -23
  m/s.

19
Average Speed

• In day-to-day usage, the terms speed and velocity
  are interchangeable.
• In physics, however, there is a clear distinction
  between these two quantities.
• The average speed of an object, a scalar
  quantity, is defined as the total distance
  traveled divided by the total time it takes to
  travel that distance
• Average Speed Total Distance/Total Time
• s d/t

20
Average Speed

• The SI unit of average speed is the same as the
  unit of average velocity meters per second.
• However, unlike average velocity, average speed
  has no direction and hence carries no algebraic
  sign.
• We can think of the total distance as the
  distance recorded by an odometer of a car.

21
Instantaneous Velocity

• Instantaneous velocity is defined as the limit of
  the average velocity as the time interval ?t
  becomes infinitesimally short.
• V ? lim(?t?0) (?x/?t)
• The notation lim(?t?0) means that the ratio ?x/?t
  is to be evaluated as the time interval ?t
  approaches 0.
• To better understand the meaning of instantaneous
  velocity as expressed by the previous equation,
  consider the following table.

22
Instantaneous Velocity

• Assume that you have been observing a runner
  racing along a track.
• One second after starting into motion, the runner
  has moved to a position 1.00 m from the starting
  point at 1.50 s, the runner is 2.25 m from the
  starting point and so on.

23
Instantaneous Velocity

• After collecting these data, you now wish to
  determine the velocity of the runner at the time
  t 1.00 s.
• The following table presents some of the
  calculations you might perform to determine the
  velocity in question.
• Thus, the average velocity in this time interval
  is 2 m/s.
• We could improve the reliability of our
  calculation by allowing the time interval to
  become even smaller, but we can state with some
  degree of confidence that the instantaneous
  velocity of the runner was 2.00 m/s at time t
  1.00 s.

24
Instantaneous Velocity

• The instantaneous speed of an object, which is a
  scalar quantity, is defined as the magnitude of
  the instantaneous velocity.
• As with average speed, instantaneous speed has no
  direction associated with it and hence carries no
  algebraic sign.

25
Graphical Interpretation of Instantaneous Velocity

• The figure below shows a position-time graph for
  a new car trip.

26
Graphical Interpretation of Instantaneous Velocity

• To find the instantaneous velocity of the car at
  point B, we must find the average velocity during
  an infinitesimally short time interval.
• This means that point C on the curve must be
  brought closer and closer to point B until the
  two points nearly overlap.
• From this construction we see that the line
  joining points B and C is approaching the line
  tangent to the curve at point B.
• The slope of the line tangent to the
  position-time curve at point B is defined to be
  the instantaneous velocity at that time.
• For the data given, the slope at point B is about
  34 m/s.
• Note that the slope at point C is zero, which
  means the car stops moving at that point.

27
Graphical Interpretation of Instantaneous Velocity

• Example
• A toy train moves slowly along a straight portion
  of track according to the graph of position
  versus time below.

28
Graphical Interpretation of Instantaneous Velocity

• Find the following
• The average velocity for the total trip.
• The average velocity during the first 4.0 seconds
  of motion.
• The average velocity during the next 4.0 seconds
  of motion.
• The instantaneous velocity at t 2.0 seconds
• The instantaneous velocity at t 5.0 seconds.
• Describe in words the trains motion from the
  origin to point A, from point A to point B, and
  from point B to point C.

29
Graphical Interpretation of Instantaneous Velocity

• The slope of the blue line joining the origin and
  point C provides the average velocity for the
  total trip.
• v ?x/?t 10 cm/12 s
• v 0.83 cm/s

30
Graphical Interpretation of Instantaneous Velocity

• B. The slope of the line joining the origin to
  point A gives us the average velocity during the
  first 4.0 seconds.
• v ?x/?t 4.0 cm/4.0 s
• v 1.0 cm/s

31
Graphical Interpretation of Instantaneous Velocity

• C. Following the same procedure for the next 4.0
  second interval, we see that the slope of the
  line between points A and B is zero.
• During this time interval, the train has remained
  at the same location, 4.0 cm from the starting
  point, and its average velocity is zero.

32
Graphical Interpretation of Instantaneous Velocity

• D. A line drawn tangent to the curve at the point
  corresponding to t 2.0 seconds has the same
  slope as the line in part B.
• Thus, the instantaneous velocity at this time is
  1.0 cm/s.
• This has to be true because the graph indicates
  that during the first 4.0 seconds of motion the
  train covers equal distances in equal intervals
  of time.
• Under these conditions the average velocity over
  any time interval and the instantaneous velocity
  at any instant are identical at all times.

33
Graphical Interpretation of Instantaneous Velocity

• E. At t 5.0 s, the slope of the position-time
  curve is zero.
• Therefore, the instantaneous velocity is zero at
  this instant.

34
Average Acceleration

• When the velocity of an object changes with time,
  the object is said to undergo an acceleration.
• Velocity of a car changes when you step on the
  gas pedal, apply the brakes, or when you alter
  your direction of motion.
• Average acceleration is the change in velocity
  divided by the time interval during which that
  change occurred
• a ?v/?ta (vf vi)/(tf ti)

35
Average Acceleration

• Example
• Suppose a car accelerates from an initial
  velocity of vi 10 m/s to a final velocity of
  vf 30 m/s in a time interval of 2.0 seconds.
  What is the average acceleration of the car?

36
Average Acceleration

• Acceleration is a vector quantity having
  dimensions of length divided by the square of
  time.
• Some common units of acceleration are meters per
  second per second (m/s/s or m/s2) and feet per
  second per second (ft/s/s or ft/s2).
• The acceleration calculated in the previous
  example was 10 m/s2, which means that on
  average, the car accelerates at a rate of 10 m/s
  every second.

37
Average Acceleration

• For the case of motion in a straight line, the
  direction of the velocity of an object and the
  direction of its acceleration are related as
  follows
• When the objects velocity and acceleration are
  in the same direction, the speed of the object
  increases with time.
• When the objects velocity and acceleration are
  in opposite directions, the speed of the object
  decreases with time.

38
Average Acceleration

• Example
• Suppose the velocity of a car changes from
  -10 m/s to -30 m/s in a time interval of 2.0
  seconds. What is the average acceleration of the
  car?

39
Average Acceleration

• The minus sign indicates that the acceleration
  vector is also in the negative x direction.
• Because the velocity and the acceleration vectors
  are in the same direction, the speed of the car
  must increase as the car moves to the left.
• NOTE The positive and negative accelerations
  specify directions relative to chosen axes, not
  speeding up or slowing down.
• The terms speeding up and slowing down refer
  to increases or decreases in speed.
• The word deceleration has a common popular
  connotation as slowing down

40
Average Acceleration

• Examples of accelerating objects
• Speed increasing direction constant.
• Speed decreasing direction constant.
• Speed constant direction changing.
• Speed increasing direction changing.
• Speed decreasing direction changing.
• Use the information below to check your answers
  to problems involving acceleration.
• v a Object is speeding up.
• v - a - Object is speeding up.
• v a - Object is slowing down.
• v - a Object is slowing down.
• v or - a 0 Object is at a constant
  velocity.
• v 0 a or - Object is speeding up from
  rest.
• v 0 a 0 Object is at rest.

41
Average Acceleration

• Quick Quiz
• Let us define eastward as negative and westward
  as positive. Answer the following statements as
  TRUE or FALSE
• If a car is traveling eastward, its acceleration
  must be eastward.
• FALSE
• 2. If a car is slowing down, its acceleration may
  be positive.
• TRUE
• 3. A particle with constant nonzero acceleration
  can never stop and stay stopped.
• TRUE

42
Instantaneous Acceleration

• In some situations, the value of the average
  acceleration differs in different time intervals.
• It is useful therefore to define instantaneous
  acceleration.
• Instantaneous Acceleration The limit of the
  average acceleration as the time interval ?t goes
  to zero.
• a ? lim(?t?0) ?v/?t
• Here again, the notation lim(?t?0) means that the
  ration of ?v/?t is to be evaluated as the time
  interval ?t approaches zero.

43
Instantaneous Acceleration

• This is a velocity-time graph for an object
  moving in a straight line.
• The slope of the blue line connecting the points
  P and Q is defined as the average acceleration in
  the time interval ?t tf ti.
• The instantaneous acceleration of an object at a
  certain time equals the slope of the
  velocity-time graph at that instant of time.

44
Instantaneous Acceleration

• Quick Quiz
• Parts a, b, and c to the right represent three
  graphs of the velocities of different objects
  moving in straight-line paths as functions of
  time.
• The possible accelerations of each object as
  functions of time are shown in parts d, e, and f.
  
• Match each velocity-time graph with the
  acceleration-time graph that best describes the
  motion.

45
Motion with Constant Acceleration

• Many of the mechanical applications discussed in
  physics will be concerned with objects moving
  with constant acceleration.
• EXAMPLE
• An object in free fall near the Earths surface
  moves in the vertical direction with constant
  acceleration, assuming that air resistance can be
  neglected.
• When an object moves with constant acceleration,
  the instantaneous acceleration at any point in a
  time interval is equal to the value of the
  average acceleration over the entire time
  interval.
• Consequently, the velocity increases or decreases
  at the same rate throughout the motion, and a
  plot of velocity versus time gives a straight
  line with either positive of negative slope.

46
Motion with Constant Acceleration

• To find the final velocity of an object moving
  with a constant acceleration after it has
  accelerated at a constant rate for any time
  interval, you can use the following equation
• vf vi a?t
• EXAMPLE
• A car starts with a velocity of 2.0 m/s to the
  right and accelerates to the right with an
  acceleration of 6.0 m/s2. What will be its
  velocity 2.0 seconds later?

47
Motion with Constant Acceleration

• To find the displacement of any object moving
  with a constant acceleration, you can use the
  following equation
• ?x ½(vi vf)?t
• EXAMPLE
• A car accelerates uniformly from rest to a
  velocity of 23.7 km/hr east in 6.5 seconds. How
  far did the car travel during this time?

48
Motion with Constant Acceleration

• Another way to find displacement with constant
  acceleration is with the following equation
• ?x vi?t ½a(?t)2
• This equation is useful not only for finding how
  far an object travels under constant acceleration
  but also for finding the distance required for an
  object to reach a certain speed or come to a stop.

49
Motion with Constant Acceleration

• EXAMPLE
• A car with an initial velocity of 23.7 km/hr west
  accelerates at a uniform rate of 0.92 m/s2 for
  3.6 seconds. Find the final speed and the
  displacement of the car during this time.

50
Motion with Constant Acceleration

• The following equation is useful in finding the
  final velocity of an object after any
  displacement
• vf2 vi2 2a?x
• EXAMPLE
• A baby sitter pushing a stroller starts from rest
  and accelerates at a rate of 0.500 m/s2. What is
  the velocity of the stroller after it has
  traveled 4.75 m?

51
Motion with Constant Acceleration

• The best way to gain confidence in the use of
  these equations is to work a number of problems.
• Many times you will discover that there is more
  than one method for solving a given problem.

52
Motion with Constant Acceleration

• PROBLEM SOLVING STRATEGY
• Make sure all the units in the problem are
  consistent.
• Choose a coordinate system.
• Make a labeled diagram of the problem including
  the direction of all displacements, velocities,
  and accelerations.
• Make a list of all the quantities given in the
  problem and a separate list of those to be
  determined.
• Think about the physical concepts involved, any
  restrictions that apply, and select from the list
  of kinematic equations the one or ones that will
  enable you to determine the unknowns.
• Check to see if your answers are reasonable and
  consistent with the diagram.

53
Motion with Constant Acceleration

• EXAMPLE
• A race car starting from rest accelerates at a
  rate of 5.00 m/s2. What is the velocity of the
  car after it traveled 100.0 ft?
• Step 1 Be sure that your units are consistent.
• We must convert 100 ft to m.
• 100 ft(1m/3.3 ft) 30.3 m

54
Motion with Constant Acceleration

• Step 2 You must choose a coordinate system. The
  origin of the coordinate system is at the initial
  location of the car, and the positive direction
  is to the right. Make a labeled diagram.

55
Motion with Constant Acceleration

• Step 3 Make a list of the quantities given in
  the problem and a separate list of those to be
  determined.
• Given
• vi 0
• a 5.00 m/s2
• ?x 30.3 m
• To Be Determined
• vf ?

56
Motion with Constant Acceleration

• Step 4 Think and select the kinematic equation
  which you will use to solve this problem.
• vf2 vi2 2a?x
• vf2 (0)2 2(5.00 m/s2)(30.3 m)
• vf2 303 m2/s2
• vf 17.407 m/s ? 17.4 m/s

57
Motion with Constant Acceleration

• You could have also solved this problem using ?x
  vi?t ½a(?t)2 to find t and then using the
  expression vf vi a?t to find the final
  velocity.
• TRY IT!

58
Freely Falling Objects

• Probably the most important situation where
  acceleration is effectively constant is
  free-fall.
• Free-fall When an object is moving vertically
  solely under the influence of gravity.
• A falling object ordinarily moves through air,
  and the essential nature of the motion is often
  obscured by fluid friction (Air Resistance).
• Nonetheless, we know that on Earth all objects
  falling through a vacuum accelerate downward at
  the same fairly constant rate, regardless of
  their weight.

59
Freely Falling Objects

• Ever since Aristotle (384 322 B.C.E.), most
  people have believed that a falling body descends
  at a rate thats proportional to its weight.
• Aristotle believed that if you had 2 stones, and
  one was 10 times heavier than the other, that the
  heavier stone would fall 10 times faster than the
  lighter stone.
• Though thats fundamentally wrong, it is a fairly
  accurate picture of the special case of extremely
  light objects, such as feathers and snowflakes,
  dropping in air.
• It applies, as well, for heavy bodies descending
  through thick fluids, like oil or honey, but it
  misses the central point.
• A stone 10 time heavier than another does not
  fall through air 10 times faster!

60
Freely Falling Objects

• The person who contributed most to our modern
  understanding is Galileo Galilei.
• Galileo really was not saying anything new when
  he asserted that, in vacuum, where there is no
  air resistance, all bodies fall at the same rate.
• By contrast, when an object falls through the
  air, moving ever more rapidly, the resistance to
  its motion, the drag, also increases.
• The greater the rate of descent, the greater the
  resistance, as more air must be pushed aside per
  second.
• Finally, a balance is reached where no further
  increase in falling speed can occur.
• This point is called terminal speed, and it
  depends on the shape, surface, and weight of the
  object.

61
Freely Falling Objects

• Example
• A sky diver, after dropping about 620 m with arms
  and legs extended spread-eagle, will reach a top
  speed of roughly 200 km/hr (120 mph).
• In a head-down dive, the same person will have a
  terminal speed up around 300 km/hr (185 mph).
• With an open parachute, drag increases
  tremendously, and the terminal speed drops to
  around 30 km/hr (20 mph).

62
Freely Falling Objects

• It was Galileo who first designed and carried out
  an experiment to confirm that the acceleration
  was constant.
• Two practical obstacles made it impossible for
  him to simply drop an object, measure vf at
  several values of t, and then compute
  acceleration for each.
• The first was that free-fall happens rapidly and
  clocks in his era were extremely crude.
• The second difficulty was that he had no way of
  measuring the instantaneous final speed at the
  end of each run.

63
Freely Falling Objects

• Here is how he solved those two problems
• First, he slowed things down instead of
  dropping balls, he rolled them along an inclined
  planes.
• He argued that if acceleration was to be constant
  for each incline, it would continue to be
  constant, even when the track was vertical and
  the ball fell freely.
• Second, he reformulated the problem, replacing vf
  by the quantities that were directly observable.
• In doing so he arrived at a statement of s vit
  1/2at2.
• If acceleration was constant, starting from rest,
  the distance traveled by the ball would depend on
  the time squared (s ? t2).
• Doubling the time a ball was allowed to move
  would quadruple the distance it traveled.

64
Freely Falling Objects

• The free-fall acceleration, due as it is to
  gravity, is represented by its own symbol, g.
• There is some variation in gravity over the
  planet, but we can generally take g to be a
  constant equal to its average value of 9.806 65
  m/s2 or 9.81 m/s2.
• In free-fall the acceleration is always g and
  its always straight downward regardless of the
  motion.

65
Freely Falling Objects

• What happens to an object if it is thrown
  vertically upward?
• Its speed continuously diminishes at a rate of
  -9.81 m/s2 as it climbs.
• The rising ball, moving slower and slower, will
  momentarily stop and then plunge downward faster
  and faster.
• The height of the stopping point, the maximum or
  peak altitude, it the vertical distance it takes
  to go from vi to vf 0.

66
Freely Falling Objects

• Ideally, the up-and-down journey is symmetrical
  in space and time around the peak altitude.
• The speed will be equal for any amount of time
  before and after the peak altitude is reached.
• The descent is exactly the same as if the object
  were dropped from the peak altitude.

67
Freely Falling Objects

• Free-Fall Equations
• vf vi - g?t
• ?y ½(vf vi)?t
• ?y vi?t ½g(?t)2
• vf2 vi2 2g?y

68
Freely Falling Objects

• Example
• A salmon is dropped by a hovering eagle. How far
  will it fall in 2.5 seconds? Ignore air drag.
• ?y vi?t ½g(?t)2
• ?y (0)(2.5 s) ½(9.81 m/s2)(2.5 s)2
• ?y - 30.656 m
• ?y ? -31 m

69
Freely Falling Objects

• Example
• A ball is thrown straight down from the roof of a
  dormitory at 10.0 m/s. If the building is 100. m
  tall, at what velocity will the ball hit the
  ground? How long will the trip take?
• vf2 vi2 2g?y
• vf2 (-10.0 m/s)2 2(9.81 m/s2)(100 m)
• vf2 -100.0 m2/s2 1962 m2/s2
• vf2 -2062 m2/s2
• vf -45.409 m/s
• vf ? -45.4 m/s
• vf vi g?t
• -45.4 m/s -10.0 m/s (9.81 m/s2)?t
• -35.4 m/s (-9.81 m/s2)?t
• ?t 3.6085 s
• ?t ? 3.61 s

70
Freely Falling Objects

• Example
• A .32-caliber bullet fired from a revolver with a
  3-inch-long barrel will have a relatively low
  muzzle velocity of about 200. m/s. If its shot
  straight up, neglecting air resistance
• What is the peak height the bullet will reach?
• How fast will it be moving when it returns to the
  height of the gun?
• How long will the whole trip take?