1. If the view you currently seeing looks like this
you are in the edit mode.
To see the full mode you need to click on the “show button:”
Located here for Powerpoint 03
Located here for Powerpoint 07 and 10

2. Notice You should now be in the full screen view
Advance to the next line by hitting the forward arrow key.
There is a lot of material in this chapter.
HOWEVER about half of it is review of material covered in the first semester of Engineering Physics.
You can click on the contents button from any slide in the presentation to go to the table of contents.

3. Clicking the Contents button on any page will return you to this page.
Kinematics Review (Review)
New Formula
Rectangular Components 3D
(Part Review)
Constant Acceleration (Review)
Projectiles (Review)
Example 1 Constant a
Example 2 Projectile
Example 3 Projectile
Example 4 Calculus
Example 5 Calculus
Example 6 Calculus
Example 7 Calculus
Normal Tangent Components
Example 8
Example 9
Example 10
Polar Coordinates
Example 11
Cylindrical Coordinates
Dependent Motion
Relative Motion
Example 12
Clicking on any of these links will take you directly to that portion of the presentation.
Clicking the Contents button on any page will return you to this page.

4. Dynamics Study of bodies that are accelerating.
A body is accelerating when:
It is speeding up or slowing down.
It is changing direction.
A combination of the above.

5. Average Velocity Time rate of change of motion
Δs = change in displacement (m or ft)
Δt = change in time (s)
v = average velocity (m/s or ft/s)

6. Instantaneous Velocity
More versatile than average
Specific speeds and directions at specific times

7. Acceleration
To analyze more complex motions you need measure of how the velocity changes.
Acceleration - time rate of change of velocity
Acceleration occurs when speeding up, slowing down, and/or changing direction.

8. Average Acceleration Δv = change in velocity (m/s or ft/s)
Δt = change in time (s)
a = acceleration (m/s2 of ft/s2)
Sign notation:
Negative acceleration DOES NOT necessarily mean slowing down

9. Sign Notation v a + + + - + 0 - + - - - 0 +/- angle What’s Happening:
Speeding up
Slowing Down
Constant
Speeding up in Negative
Turning

10. General Rule v and a same sign v and a opposite sign v and a at angles
speeding up
slowing down
turning

11. Instantaneous Acceleration
More versatile than average
Specific accelerations and directions at specific times

12. Short Definitions Time measures Displacement Velocity Acceleration
When
Where
How fast and which way
How velocity changes
(Speeding up, slowing down and/or changing direction)

13. General Case curvilinear motion
Given an equation for x or v or a with respect to time.
Use calculus to determine the other equations.

14. Another equation. Combine the velocity and acceleration equations:

15. Curvilinear motion Rectangular Components
When the position of a particle is given in terms of x, y and z components. Its dynamics can be analyzed by use of unit vectors and calculus

16. Curvilinear motion Rectangular Components

17. A word on notation
There are three ways to indicate that you must take the derivative of an equation.
Where s is some equation with t as its variable s(t)
A second derivative may be indicated by:
The book uses the dot notation. The equation writer I use for these programs does not support the dot format. So all future notation will be of the prime format.

18. Special Case 1: Constant Acceleration
Case where a = some constant value
Derive some useful equations
Less need for calculus

23. Remember a is constant

40. Constant Acceleration Equations also called Derived Equations

41. Special Case 2: Projectiles
Objects moving in two dimensions.
Gravity is the only force acting.

42. Symmetry of Motion Path is parabolic
Ball thrown at an angle above the horizon with no air friction.
Path is parabolic

43. Symmetry of Motion
Ball thrown at an angle above the horizon with no air friction.
Velocities:
(1) Where is the minimum velocity?
Top
(2) What is the minimum velocity?
The answer is obvious if we break all of the velocities into their X and Y components.

44. Symmetry of Motion
Ball thrown at an angle above the horizon with no air friction.
Velocities:
(1) Where is the minimum velocity?
Top
(2) What is the minimum velocity?
Top = The initial X component
(3) Where is the maximum velocity?
At the lowest point(s) in the flight
(4) Where are some equal velocity pairs?
All x components are equal. y’s are in opposite direction so are not equal.
5) Where are some equal speed pairs?
At equal heights above the ground.

45. Projectiles - Two Dimensions
X axis constant velocity
(assuming air friction minimal!)
Y axis constant acceleration equations
Add two motions together. (vector addition)

46. Formulas (nothing new)
X axis motion constant velocity
vx = vT cos x = vxt
Y axis motion derived equations
voy = vT sin y = voyt - ½ gt2
Add two motions together.

47. When to use what Ask a two questions:
1. Is object moving at a constant velocity?
Yes - Constant velocity equation
No - Go to next question
2. Is object moving at a constant acceleration?
Yes - Use constant acceleration equations
No - Use calculus

48. Clues
Clues that will tell you that the velocity of the object is constant:
v = 4 m/s (or any other number)
“a constant rate” or “a constant rate of 4 m/s”
a = 0 (acceleration is zero)
x = 4t (an equation for displacement to the first power. Notice that the derivation of x, the velocity, is a constant)

49. Clues
Clues that will tell you that the acceleration of the object is constant:
a = 4 m/s2 (or any other number)
“an acceleration of 4 m/s2”
a = 0 (acceleration is zero)
x = 4t2 (an equation for displacement to the second power. Notice that the second derivation of x, the acceleration, is a constant)
v = 4t (an equation for velocity to the first power. Notice that the derivation of v, the acceleration, is a constant)
“an object is going in a curved path with a constant velocity”
“an object is going in a circular path with a constant rotation”

50. Clues
Clues that will tell you that the acceleration of the object is non constant:
a = 4t (or any other equation)
x = 4t3 (an equation for displacement to the third power or higher . Notice that the second derivation of x, the acceleration, is not constant)
v = 4t2 (an equation for velocity to the second power or higher. Notice that the derivation of v, the acceleration, is a non constant)
an object is going around any NON circular path at a constant velocity
In general if you see an equation be suspicious that you need to use calculus.

51. Clues
Clues that will tell you that the object is in free fall - a projectile
An object is thrown, shot, catapulted (etc)
Any situation where the force is exerted then the object is released with the force no longer present.
Must be in a gravitational field - a ball thrown on Mars or the Moon is a projectile a ball thrown in free space is not.
The object must be something that does not have an internal source of energy or airfoils - no engines, parachutes, wings. An acorn is a projectile, a dandelion seed is not a projectile.

52. Consider the following multiple part situation
A plane loads passengers, races down the runway until it reaches a speed of 200 MPH. At that speed it then gains altitude and banks in a circle such that it changes direction to the East. Once it arrives at its cruising altitude of 5000 ft it travels at a constant rate for the next 2 hr. At its destination it descends following a parabolic path until it touches the end of the runway. It then decelerates at 30 ft/s2 until it comes to a stop.

53. What type of equation would you use for each part?
1. A plane loads passengers, races down the runway until it reaches a speed of 200 MPH.
What type equation would you use for this part of the motion? Why? Choices: constant v, constant a, calculus, not enough information. Make a choice now THEN hit the advance key.
Actually you don’t know, there is not enough information. Typically a plane takes off with a constant thrust. This means that the acceleration will actually increase, especially with large jet planes, because the fuel is being burned and the planes mass is decreasing. This means you must use calculus. HOWEVER a small plane does not burn as much fuel so a constant acceleration approach would work

54. What type of equation would you use for each part?
2. At that speed it then gains altitude and banks in a circle such that it changes direction to the East.
What type equation would you use for this part of the motion? Why? Choices: constant v, constant a, calculus, not enough information. Make a choice now THEN hit the advance key.
Constant acceleration. The first three words tell you that the speed does not change BUT the rest of the sentence tells you that the direction does and it is a circular path. This means that there is an acceleration (the good old centripetal acceleration)

55. What type of equation would you use for each part?
3. Once it arrives at its cruising altitude of 5000 ft it travels East at a constant rate for the next 2 hr.
What type equation would you use for this part of the motion? Why? Choices: constant v, constant a, calculus, not enough information. Make a choice now THEN hit the advance key.
Constant velocity. The speed does not change, the direction does not change

56. What type of equation would you use for each part?
4. At its destination it descends following a parabolic path until it touches the end of the runway.
What type equation would you use for this part of the motion? Why? Choices: constant v, constant a, calculus, not enough information. Make a choice now THEN hit the advance key.
Calculus. The plane changes direction that tells you that there is an acceleration. The curve is non circular that tells you that the acceleration is not constant.
We assume that the curve is such that the plane is traveling parallel to the ground just before it hits the ground.

57. What type of equation would you use for each part?
5. It then decelerates at 30 ft/s2 until it comes to a stop.
What type equation would you use for this part of the motion? Why? Choices: constant v, constant a, calculus, not enough information. Make a choice now THEN hit the advance key.
Constant acceleration. The number and dimensions tell you.

58. Example 1: In a speed trap, two pressure activated strips are placed 110 m apart across a highway on which the speed limit is 90 km/hr. This device measures the average velocity of a car. While going 120 km/hr, a driver notices a police car and manages to hit the brakes to slow down just as he activates the first strip. What constant deceleration is needed so that the car’s average speed is within the speed limit when the car crosses the second marker?

59. First we need consistent dimensions:
v0 = 120 km/hr = m/s
vAve = 90 km/hr = 25 m/s
An average is the sum of the cases divided by the number of the cases so:
vAve = (v0+v)/2
25.0= (33.33+v)/2
v=17 m/s

60. This is a constant acceleration problem so we can supply the appropriate equation:
Since the typical deceleration of a car is on the order of -10 m/s2 this is certainly possible.
Although the policeman would probably give you a ticket for being a smart aleck

61. Example 2: A ball is thrown at an angle of 35o with a speed of 27 m/s
Example 2: A ball is thrown at an angle of 35o with a speed of 27 m/s. It lands on top of a 7.5 m tall building. What is the total velocity of the ball just before it lands?
Drawing: x y
Numbers:
vx=(27m/s)cos35 = 22.1m/s
v0y=(27m/s)sin35=15.5m/s
vx= 22.1m/s
v0y= 15.5m/s
a = -9.8m/s2
y = 7.5m

62. Identify an equation: Solve the equation:
From the drawing the ball must be going down when it hits the top of the building. So we discard the positive answer.

63. Evaluate: The units and size are correct. This is not the answer. Why?
The question asked for the total velocity. We have the final y velocity and the x velocity (remember the x does not change)
Since velocities are vectors we must use vector addition.
We already have the x and y parts of the answer so we go directly to Pythagorean’s theorem.

64. Remember the magnitude of a vector is always positive
Remember the magnitude of a vector is always positive. Direction is indicated by the angle.
So we need to calculate the angle:

65. Now find the standard position angle.
A minus y and a plus x places it in the fourth quadrant so:

66. The final answer is: 24.1 m/s336o
Evaluation: The dimensions are correct.
The angle is in the fourth quadrant which is correct.
The size is less than the original total velocity. This makes sense because the ball ended up above the starting point.

67. Example 3: A ball is thrown from the top of a building with a velocity of 25 m/s at an angle of 35 degrees below the horizontal. If the ball is 45 m above the ground when released (a) how long does it take for the ball to hit the ground? (b) how far from the base of the building along the ground is the ball at that point?

68. Find the initial velocities:

69. We know three things about the y motion so solve for time:

70. Finally we can solve for x

71. 4. A dirt bike can accelerate according to a =
4. A dirt bike can accelerate according to a = .1t2 A stunt man riding the bike needs to cross a 5 m wide ditch. The side the bike starts on is horizontal. The landing side is 3 m lower than the take off side. The wheelbase of his bike is 1.15 m. In order to make the jump impressive he wants to land with his rear wheel on the edge of the far side of the ditch. (a) If he makes a standing start, how far from the ditch on the take off side must the bike start? (b) What velocity must the bike have when it leaves the take off side?
Known: acceleration part jump part

72. We can find equations for the velocity and displacement of the bike on the take off side by the use of calculus
This helps some but not a lot because we have no times.
So we have to consider the jump part of the problem.

73. The jump is a projectile motion with a horizontal initial angle
The jump is a projectile motion with a horizontal initial angle. This means:
The jump starts when the front wheel leaves the ground and ends when the rear wheel hits the ground on the far side. So we need to include the wheelbase of the bike:
Known: acceleration part jump part

74. The jump is a projectile motion with a horizontal initial angle
The jump is a projectile motion with a horizontal initial angle. This means:
The jump starts when the front wheel leaves the ground and ends when the rear wheel hits the ground on the far side. So we need to include the wheelbase of the bike:
Known: acceleration part jump part

75. Now we can find the time of the jump with the Y information:
Next the constant X velocity during the jump:
Which is the answer to part (b)
Known: acceleration part jump part

76. We can now return to the acceleration part of the problem:
Finally the distance:
Which is the answer to part (a)
Known: acceleration part jump part

77. Example 5: A robot’s hand moves according to VX = 3m/s vY = (
Example 5: A robot’s hand moves according to VX = 3m/s vY = (.25t)m/s and vZ = (-3sin(2πt))m/s. At t = 0 and velocities are zero. (a) What is the hands displacement at t = 2.5 s in rectangular format? (b) What is the hands acceleration at t = 2.5 s in rectangular format?
Inspection reveals that the x velocity is constant, the y motion has a constant acceleration (the v equation involves t to the first power) and the z motion has a variable acceleration.
We could use three different techniques – but it is easier to do the whole problem with calculus.

78. First integrate to find the three displacement equations and evaluate at t = 2.5s
State the answer in rectangular format:

79. Next derive to find the three acceleration equations and evaluate at t = 2.5s
State the answer in rectangular format:

80. Example 6: The location of a rocket at any time t while the motor is running is given by the equations:
y = 5(t-5) x = 3t
Where y is the height in m, x is the horizontal range in m and t is the time in s. (a) What is the acceleration of the projectile at t = 2.3 s? (b) After 7 s the motor shuts off and the rocket falls to the ground. How long after the rocket takes off does the rocket hit the ground?

81. In the Y this is a case of a non-constant acceleration and we must use calculus:

82. In the X this is a case of a constant velocity.
Hint: if you do not recognize a case (constant v, constant a, non-constant a) calculus will always work although it may not be easy. For example in this case:
The fact that the acceleration in the x direction is zero tells you that the velocity in the x direction is constant.

83. A word about dimensions:
I have a habit (most people do) of omitting the dimensions from the calculations.
The actual equations, with dimensions, should be:
y = (5m/s3)(t-5s)3+(625m) x = (3m/s)t
When you put a time in seconds in the equations they give correct dimensions for x and y which is meters.

84. When you use any mathematical process on an equation with dimensions, treat the dimensions as a constant.
For example the derivations with the dimensions should look like:
When you put a time in seconds into either equation it gives the correct dimensions v in m/s and a in m/s2

85. My tendency is to leave the dimensions out of the problem and put the correct dimensions on the answer.
This works OK as long as I have the correct formula and put the correct dimensions on the answer.
But I will admit it is a sloppy habit.

86. Back to the problem:
ax = 0 so the total acceleration is ay
so the answer to part (a) must be:

87. In the second part the rocket runs out of fuel, the motor shuts off and it goes into freefall.
We know that the acceleration is 9.8 m/s2 down.
We need to know the position and velocity of the rocket when the fuel runs out. These are the initial conditions of the free fall part of the motion
We find these from our previously derived equations

88. At t = 7 s we have:
In the Y a projectile has a constant acceleration and in the X a projectile has a constant velocity.

89. voy = 60 m/s ay = g = -9.81 m/s2 y = -665 m
Look at the Y part of the projectiles free fall. The known measures are:
voy = 60 m/s ay = g = m/s2 y = -665 m
Note: The y is negative because the rocket starts out up in the air and ends up 665 m below its starting point.
Note: the negative time is discarded since it means that the rocket got to -665 m before the engine stopped.

90. Finally to find the total time in the air add the two times for the powered and projectile flights:
tT = = 26.3s
While we have additional information about the x times, and velocities we don’t need them. So we drop them.

91. Example 7: The engine of a rocket delivers a constant thrust
Example 7: The engine of a rocket delivers a constant thrust. At the same time the engine uses fuel at a high rate. This means that the thrust (force) is constant and the mass decreases or from F=ma the acceleration must increase. In turn this means that the acceleration changes as the displacement changes. Given the acceleration displacement curve shown calculate the velocity of the rocket when it has traveled 200 ft. How long does it take for the rocket to travel 200 ft?
a=3s2
displacement
acceleration

92. Hint: One way to use the equation ads = vdv is to find an equation for a in terms of s

93. Hint: One trick used in problem solving is to start with a math definition which is a derivative, rearrange it, then integrate. This trick can be used to solve part b.

94. Notice: The previous material (except for the final example) was review and should have been covered in the first semester Physics course. Additional explanations and help can be had by reviewing the physics text - assuming you have not sold it.
“Now for something completely different.”

95. Curvilinear motion normal and tangential components
When the path of the particle is known coordinates of tangent to and normal to the path can be useful.
For example a car driving a country road may go through a series of curves to the left and right and up and down over hills and through valleys.
Using Cartesian XY equations to deal with the force that causes such a motion can become quite complicated.
Using normal tangential equations can be less complicated.

96. In such a motion the total force moving the car constantly changes in magnitude and direction. The force will always be toward the center of curvature and may be toward the direction of motion (if the car is turning and speeding up)
or inward and against the direction of motion (if the car is turning and slowing)
or directly toward the center of curvature (if the car is turning with a constant speed.
and so on

97. Using the tangential, normal system breaks those vectors into components.
The effect of the force on the speed of the vehicle (the tangent components green) is clearly seen
The effect of the force on the direction of the vehicle (the normal components blue) is also seen.
In addition the two motions can now be dealt with separately.

98. Sign Notation Velocity
Positive tangent axis is in the direction of motion.
Positive normal axis is to the center of curvature.
Velocity
Direction of the velocity is always tangent to the path.
Magnitude is found by using the first derivative of the path equation.
You must have an equation in terms of s displacement along the path from the starting point and t time

99. Normal and tangential components of Acceleration
Causes a change in the speed.
Positive in the direction of motion.
Magnitude is found by the standard techniques:

100. Normal and tangential components of Acceleration
Causes a change in the direction.
Positive towards the center of curvature.
Magnitude is found by:
Where  is the radius of curvature found by:
If the path is a circle
 = radius of circle
In all other cases radius is

101. Normal and tangential components of Acceleration
Three Dimensional
We can make the normal tangential co ordinate system into a three dimensional system by adding a third axis perpendicular to the other two.
This axis is labeled the “b” axis and the properties are the same as any rectilinear axis.
That is: displacement on the b axis is an equation of distance as a function of time. s(t) = b
Velocity is the first derivative of s vb = ds/dt
Acceleration is the first derivative of velocity ab=dvb/dt

102. Example 8. A car at an Indy type race accelerates uniformly from the pit area, going from rest to 320 km/hr in a semicircular (A to B to C) arc with a diameter of 450 m. Determine the total acceleration of the car when it is halfway through the turn
Drawing:
Known:
Convert appropriate numbers to standard units
d = 450m C A B
vA = 0
vC =320 km/hr
r = 225 m

103. The angle traveled is 180o so:
Calculate the tangential acceleration at point B. Much of this problem could be worked using linear equations but lets convert everything to circular format.
The angle traveled is 180o so:
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m

104. Solve the equation then continue to check.
Since the tangential acceleration is constant we can find it using the constant acceleration equations:
Solve the equation then continue to check.
This acceleration is constant throughout the curve so now we can find the velocity at B.
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m
ωA = 0
ωC = .395 rad/s
θ = 3.14 rad

105. We want the velocity at point B so the angle is ¼ of a circle
Solve the equation then continue to check.
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m
ωA = 0
ωC = .395 rad/s
θ = 3.14 rad
α = .0248rad/s2

106. The tangential acceleration is the linear form of the angular acceleration (α) calculate the acceleration then continue:
To find the normal (centripetal) acceleration we need the linear velocity at B. Calculate that velocity then continue to check:
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m
ωA = 0
ωC = .395 rad/s
θ = 3.14 rad
α = .0248rad/s2
ωB = .279 rad/s

107. Since this is a circular path the radius of curvature is the radius of the circle. OR we can show this is true (I used to do the derivations)

108. Calculate the normal acceleration at B then continue:
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m
ωA = 0
ωC = .395 rad/s
θ = 3.14 rad
α = .0248rad/s2
ωB = .279 rad/s
aTan = 5.58m/s2
aBN = 17.5m/s2
vB = 62.8m/s

109. Find the total acceleration:
Evaluate: A standard car accelerates at between 6 and 10 m/s2 so this acceleration for an Indy car is believable.
d = 450m C A B
vA = 0
vC = 88.9 m/s
r = 225 m
ωA = 0
ωC = .395 rad/s
θ = 3.14 rad
α = .0248rad/s2
ωB = .279 rad/s
aTan = 5.58m/s2
aCen = 17.5m/s2
vB = 62.8m/s

110. Example 9: The most common curve for horizontal changes in direction of roadways (going around a corner) is the circle. The most common curve for vertical changes (over hills etc.) is parabolic. As a road goes over a hill it follows the equation y = -6.5x10-5x2 +.04x. As a truck goes up hill (from x = 0 to x = 325 ft) it slows at the rate of 10 ft/s2 to 45 ft/s at the top of the hill. After it goes over the top of the hill it speeds up at the rate of 18 ft/s2 until it reaches the bottom of the curve going at 91 ft/s. (a) What is the total acceleration at the top of the hill (x = 325 ft)? (b) What is the total acceleration at the bottom (x = 500ft)?

111. Calculate the radius of curvature at x = 325 ft

113. (b) Calculate the radius of curvature at x = 500 ft

115. 200 mm
100 mm
Example 10: when a bead reaches the point (200,100) mm on the wire shown its speed is 300 mm/s and is increasing at a rate of 225 mm/s2. The curve is defined by y = (20)(103)/x Determine the total acceleration at that point

117. Now add the normal and tangential components together to get:
path
acceleration
34.9o

118. Polar Co-ordinates
When the path equation is given in terms of an angle and a distance to the particle (polar co-ordinates)
Radial component (r) is established by a line from the origin directly to the particle.
Transverse component () is measured counterclockwise from a fixed line (often +x axis) to the radial line.

119. and the angle made with the +X axis (θ) change.
Polar co-ordinates
As a particle moves along a curved path the distance to the origin (r) shown in red.
and the angle made with the +X axis (θ) change. θ θ θ

120. Polar Equations
Polar equations r and θ can be written in a couple of different formats:
Notice that the second case, on the right, will take some special handling when taking derivatives.

121. Using Polar components to calculate the velocity of a body
Velocity: Components
Velocity: Magnitude
Direction is tangent to the path.

122. Using polar notation to calculate the acceleration of the body
Acceleration: Components
Acceleration: Magnitude
Direction can be found by using the inverse tangent function.

123. Example 11: A particle travels along a path defined by r = e5t cm and θ = 3t2 rad/s Determine the magnitudes of the total velocity and total acceleration when t=.25 s

124. First lets calculate the first and second derivatives of the functions and evaluate them at t = .25 s

125. Next we will apply the evaluated formulas to the appropriate equations

126. Now to really mix things up
You can make the polar system into a three dimensional system by adding a z component.
In this case the z component takes the shape of an equation for z in terms of time, just like a three dimensional rectangular system.
When you do this you have a Cylindrical Coordinate system.

127. Cylindrical Coordinate system
Velocity: Components
Velocity: Magnitude
Direction is tangent to the path.

128. Cylindrical Co-ordinates
Acceleration: Components
Acceleration: Magnitude
Direction can be found by using the inverse trig functions - you must have 2 angles azimuth and elevation. (See your statics book.)

129. Dependent Motion
In some cases, primarily involving pulley systems, the motion of one object is dependent upon the motion of another object.
In such a case we can write an equation which expresses the relation between the two objects.

130. Consider the pulley system shown. A B
Consider the pulley system shown.
The lines shown are actually a continuous rope.
This means as A moves down B moves up BUT not as far or as fast.
By writing the dependent equations we can determine the relation.

131. A B
1. Draw the position co-ordinates. Basically these are lines through the axis of each pulley perpendicular to the lines that lead away from that pulley.
2. The position co-ordinates have broken the line into pieces. Label each piece.
In this case we have two segments labeled sB. These segments are the same length AND will remain the same length as B moves. sA sB

132. 3. Write an equation which gives the length of the line
3. Write an equation which gives the length of the line. Basically add all of the line segments together.
l = sA + 2sB
4. Take the first derivative of the line equation to get the velocity equation:
0 = vA + 2vB
vA = - 2vB
Which makes sense. As A moves down B moves up. When A moves 2 m B moves 1 m (because of the two ropes supporting B) A B sA sB

133. 5. Take the first derivative of the velocity equation to get the acceleration equation:
0 = aA + 2aB
aA = - 2aB A B sA sB

134. Relative motion
If you have ever taken a long trip in a car at a constant velocity, say 60 mph, you may have experienced the feeling that you were standing still and the landscape was moving past you at 60 mph.
Further if you observed a car passing you, when you feel you are not moving, it seems to be moving very slow say 5 mph, instead of its speedometers reading of 65 mph.
Finally cars coming at you seemed to be moving very fast, 120 mph, while you are standing still

135. Relative motion
All of the previous cases are examples of relative velocities
In the second case the passing car has a velocity of 65 mph relative to the ground. You have a velocity of 60 mph relative to the ground. But when you are passed the passing car has a speed of only 5 mph relative to you.
In the third case both you and the oncoming car have velocities of 60 mph relative to the ground but the oncoming car has a velocity of 120 mph relative to you.

136. Relative motion
It is often useful to find the motion of one particle in relation to a second particle.
This is done by finding the vector quantity of both particles relative to a fixed position and solving the vector equation:
The notation rB/A stands for r of B relative to A

137. Example 12: Car 1 is moving at a constant velocity of 20 ft/s 30o standard position relative to the ground. Car 2 is moving at 50 ft/s 210o standard position angle relative to the ground. What is the velocity of the car 2 relative to car 1? What is the velocity of the car 1 relative to car 2?
In the relativity formula vB/A stands for velocity of B with respect to A
We want velocity of car 2 relative to car 1
Therefore vB = v2 = 50 ft/s at 210o
vA = v1 = 20 ft/s at 30o

139. Part b asks for the velocity of car 1 relative to car 2.
You could go through the whole process again.
Or you could think about the situation and realize that this relative velocity must be exactly the same magnitude with exactly the opposite direction.
So v1/2 = 70 ft/s at 30o

140. Explanation of the explanation
Consider the case where you are in a car (2) traveling at 60 mph north while another car (1) approaches you at 60 mph south.
To you the other car has a relative velocity of v1/2 = 120 mph south.
To the other car you have a relative velocity of v2/1 = 120 mph north.
So when you switch relative positions the direction of the velocity reverses.

141. For the Test
You will be expected to be able to solve a kinematics problem involving more than one type of motion (constant velocity, constant acceleration, non constant acceleration) You will need to be able to recognize when to use calculus and when to use the acceleration formulas.
You should know the normal tangent coordinate system well enough to solve problems in that system.
You should be able to analyze pulley systems.