Motion in a Straight Line

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Motion in a Straight Line
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KINEMATICS

• - the process of motion is integral to the
  description of matter characteristics
• - all matter is moving - therefore a method must
  be formulated for accuracy

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DISTANCE vs. DISPLACEMENT

• 1. DISTANCE - defined as the magnitude or length
  of motion - NO DIRECTION INDICATED symbol
  d
• 2. DISPLACEMENT - magnitude and direction of
  motion symbol s

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Measurement of Speed

• Total distance the sum of the all changes in
  position
• Time an interval of change measured in seconds
• Position Separation between the object and the
  reference point

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Types of Speed

• Rest no change in motion
• Instantaneous the current speed of an object at
  a point of time
• Average two ways to determine the mean
  movement
• Total distance / Total time
• Sum of the individual speeds / number of speed
  measurements
• Speed is a scalar quantity

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Velocity

• Displacement the change in position based on
  distance and Direction
• Velocity is the change in Distance per unit
  time with a specific direction
• Velocity is a vector quantity
• Velocity also has the following conditions
• Average
• Instantaneous

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Constant Speed

• An object could be moving at a steady rate.

Thus, its average and instantaneous speed would
be the same!
GBS Physics - speed vs. velocity
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Question

• A car is traveling at a constant 60mi/hr in a
  circular path.

Does it have a constant speed? Does it have a
constant velocity?
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Problem Solving

• G write down what information is given (with
  units)
• U Identify what you have to find the unknown
  (with units)
• E Identify an equation (relationship) that
  equates the givens and unknown
• S Solve the equation for the unknown
  (algebraically before substituting any of the
  "givens" with their units because there will be
  fewer mistakes in copying the units until the
  last step).
• S Substitute the givens (with units) and find
  the answer. 

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Sample Problem 1

• If a car is traveling at an average speed of 60
  kilometers per hour, how long does it take to
  travel 12 kilometers?
• GIVEN Ave Speed 60 km/hr
• Distance 12 km
• UNKNOWN Time - ?
• EQUATION v d/t

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Sample Problem 1

• If a car is traveling at an average speed of 60
  kilometers per hour, how long does it take to
  travel 12 kilometers?
• SOLVE v d/t gt t d/v
• SUBSTITUTION t 12km / 60km/h
• t .2 h
• t 12 min
• t 720 sec

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Sample Problem 2

• A high speed train travels 454 km in 7200
  seconds. What is the trains average speed?
• GIVEN Time 7200 s (always a good idea to
  convert to (base / fundamental units) - time to
  seconds- if needed!)
• Distance 454 km - (always a good idea
  to convert to (base / fundamental units) - km to
  meters- if needed!)
• 454 km 454000 m
• UNKNOWN Ave Speed - ?
• EQUATION v d/t

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Sample Problem 2

• A high speed train travels 454 km in 7200
  seconds. What is the trains average speed?
• SOLVE v d/t gt v d/t
• SUBSTITUTION v 454km / 7200s
• (with units!) v 454000m / 7200s
• v 63 m/s

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Example

• If sun light takes about 8 minutes to go from the
  sun to the earth, how far away from the sun is
  the earth? Hint light travels at 186,000
  miles per second!!!
• GIVEN Time 8 min (always a good idea to
  convert to (base / fundamental units) - time to
  seconds- if needed!) 8 min 480 s
• Ave Speed 186,000 mi/s
• UNKNOWN Distance - ?
• EQUATION v d/t

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Example

• If sun light takes about 8 minutes to go from the
  sun to the earth, how far away from the sun is
  the earth? Hint light travels at 186,000
  miles per second!!!
• SOLVE v d/t gt d vt
• SUBSTITUTION d 186,000 mi/s X 480s
• (with units!) d 89,000,000mi

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Summary

• Speed is based on position change relative to the
  origin
• Scalar and Vectors quantities are used to
  describe motion
• Calculations must include formula, substitution
  of proper units, and final solution in the MKS
  system.

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Relativity of Velocity

• Theory, developed in the early 20th century,
  which originally attempted to account for certain
  anomalies in the concept of relative motion, but
  which in its ramifications has developed into one
  of the most important basic concepts in physical
  science
• Velocity changes when compared to a frame of
  reference

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Acceleration

• Acceleration rate of change of velocity
• Acceleration describes how fast an objects speed
  is changing per
• amount of time.

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Types of Acceleration

• Acceleration , also known as linear acceleration,
  rate at which the velocity of an object changes
  per unit of time. A Dv/t (Average
  Acceleration)
• Uniform Acceleration the constant rate of
  change in Velocity ( Free Fall )
• 9.81 m/s2 (use 10 m/s2 in multiple choice)

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• If an object has a constant velocity, then its
  acceleration would be zero.

If an object is slowing down, it is decelerating
or a NEGATIVE acceleration
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Formulas
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How to choose the best formula

• Free Fall
• Acceleration due to gravity
• Uniform acceleration
• Distance is not part of the question
• Time is part of the question

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How to choose the best formula

• Free Fall
• Acceleration due to gravity
• Uniform acceleration
• Distance is part of the question
• Time is part of the question

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How to choose the best formula
Choose this formula when the question does not
include the TIME
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HW p. 2 Q1 - a

• A 60 mi/hr wind is blowing toward the S. What is
  the resultant velocity of an airplane traveling
  100 mi/hr when it is heading

Resultant 40 mi/hr N
Plane - 100 mi/hr N
Wind - 60 mi/hr S
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HW p. 2 Q1 - b

• A 60 mi/hr wind is blowing toward the S. What is
  the resultant velocity of an airplane traveling
  100 mi/hr when it is heading

Wind - 60 mi/hr S
Plane - 100 mi/hr S
Resultant 160 mi/hr S
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HW p. 2 Q1 - c

• A 60 mi/hr wind is blowing toward the S. What is
  the resultant velocity of an airplane traveling
  100 mi/hr when it is heading

Wind - 60 mi/hr S
Resultant 116.6 mi/hr _at_
S 59º E
Plane - 100 mi/hr S
a2 b2 c2 (60 mi/hr)2 (100 mi/hr) 2 c2 C
116.6 mi/hr ? tan -1 (100 mi/hr / 60 mi/hr)
59º
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HW p. 2 Q1 - c

• A 60 mi/hr wind is blowing toward the S. What is
  the resultant velocity of an airplane traveling
  100 mi/hr when it is heading
• GIVEN a. v 40 mi/hr b. v 160 mi/hr c. v
  116.6 mi/hr t 5 hr
• UNKNOWN d - ?
• EQUATION v d/t
• SOLVE v d/t gt d vt
• SUBSTITUTION a. d (40 mi/hr)(5 hr) 200 mi
• (with units!) b. d (160 mi/hr)(5 hr) 800
  mi
• c. d (116.6
  mi/hr)(5 hr) 583 mi

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HW p. 2 Q2

• Rowboat across a stream flowing _at_ 3 mi/hr. Boy
  can row boat _at_ 4 mi/hr directly across stream.

Boat - 4 mi/hr
Water 3 mi/hr
Resultant 5 mi/hr _at_
36.9º
a2 b2 c2 (3 mi/hr)2 (4 mi/hr) 2 c2 C 5
mi/hr ? tan -1 (3 mi/hr / 4 mi/hr) 36.9º
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HW p. 3 Q1

• A fish swims at the rate of 2 ft/s. How long will
  it take this fish to swim 36 ft?
• GIVEN Ave speed 2 ft/s d 36 ft
  
• UNKNOWN time - ?
• EQUATION v d/t
• SOLVE v d/t gt t d/v
• SUBSTITUTION t 36 ft / 2 ft/s
• (with units!)
• t 18 s

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HW p. 3 Q3 part a

• A car starts from rest accelerates up to a
  velocity of 40 ft/s in 10 s?
• GIVEN vi 0 ft/s vf 40 ft/s t 10 s
  
• UNKNOWN a - ?
• EQUATION vf vi at
• SOLVE vf vi at gt vf - vi at
• vf vi / t a
• SUBSTITUTION a vf vi / t (with units!)
• a (40 ft/s O ft/s) / 10 s 4 ft/s2

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HW p. 3 Q3 part b

• A car starts from rest accelerates up to a
  velocity of 40 ft/s in 10 s?
• GIVEN vi 0 ft/s vf 40 ft/s t 10 s
  
• UNKNOWN Ave speed ?
• EQUATION Ave speed vf vi / 2
• SOLVE ave V vf vi / 2
• SUBSTITUTION ave V vf vi /2
• ave V (40 ft/s 0 ft/s) / 2 20 ft/s

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HW p. 3 Q3 part c

• A car starts from rest accelerates up to a
  velocity of 40 ft/s in 10 s?
• GIVEN vi 0 ft/s vf 40 ft/s t 10 s a 4
  ft/s2 ave v 20 ft/s
• UNKNOWN d ?
• EQUATION Ave v d/t
• SOLVE Ave v d/t gt d (ave v)t
• SUBSTITUTION d (ave v)t
• (20 ft/s)10 s 200 ft

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Sample Problem 1

• A brick falls freely from a high scaffold at a
  construction site.
• What is the velocity after 4 seconds?
• How far does the brick fall in this time?

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Solution Given a 9.8 m/s2 t 4s
What is the velocity after 4 seconds? Find V
Vf 0 m/s (-9.8 m/s2) ( 4.0 s) -39.2 m/s
How far does the brick fall in this time? Find d
d 0 m/s (4s) .5(-9.8 m/s2) (4s)2 0
.5(-9.8m/s2) (16 s2 ) -78.4m
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Sample problem 2

• An airplane must reach a speed of 71m/s for
  takeoff. If the runway is 1000m long, what must
  be the acceleration?

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Solution

• What is the acceleration needed to take off?

Given Vi0 m/s Find a ? Vf71 m/s d1000m
(71m/s)2 (0 m/s)2 2 (a) (
1000m) (-2000m) a - 5041 m2 / s2
a 2.5 m/s 2
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Summary

• Determine the type of motion
• List the given information
• Choose the best formula from the Physics
  formulas
• Substitute the proper units
• Solve for the unknown in the equation

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GRAPHICAL REPRESENTATION OF VELOCITY

• slope - the slope of a displacement vs. time
  curve would be the velocity

GBS Physics - position vs. time
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Constant Velocity Positive Velocity
Positive Velocity Changing Velocity
(acceleration)
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Slow, Rightward () Constant Velocity
Fast, Rightward () Constant Velocity
http//www.physicsclassroom.com/mmedia/kinema/cpv.
html
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Slow, Leftward (-) Constant Velocity
Fast, Leftward (-) Constant Velocity
http//www.physicsclassroom.com/mmedia/kinema/cnv.
html
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GRAPHICAL REPRESENTATION OF ACCELERATION

• slope - the slope of a velocity vs. time curve
  would be the acceleration

http//www.glenbrook.k12.il.us/gbssci/phys/mmedia/
kinema/avd.html
GBS Physics - velocity vs time
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Positive Velocity Positive Acceleration
Positive Velocity Zero Acceleration
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Determining the Area on a v-t Graph
As previously learned, a plot of velocity-time
can be used to determine the acceleration of an
object (the slope). We will now learn how a plot
of velocity versus time can also be used to
determine the displacement of an object. For
velocity versus time graphs, the area bound by
the line and the axes represents the
displacement.
http//www.glenbrook.k12.il.us/gbssci/phys/Class/1
DKin/U1L4e.html
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The shaded area is representative of the
displacement during from 0 seconds to 6 seconds.
This area takes on the shape of a rectangle can
be calculated using the appropriate equation.
Area b h Area (6 s) (30 m/s) Area 180
m
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The shaded area is representative of the
displacement during from 0 seconds to 4 seconds.
This area takes on the shape of a triangle can be
calculated using the appropriate equation.
Area 0.5 b h Area (0.5) (4 s) (40
m/s) Area 80 m
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The shaded area is representative of the
displacement during from 2 seconds to 5 seconds.
This area takes on the shape of a trapezoid can
be calculated using the appropriate equation.
Area 0.5 b (h1 h2) Area (0.5) (3 s)
(20 m/s 50 m/s) Area 105 m
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