Chapter

Vectors and Two-Dimensional Kinematics

Blinn College - Physics 2325 - Terry Honan

. - Vector Algebra - I

Polar Coordinates

(x,y) are the Cartesian (or rectangular) coordinates of some point on a plane.  r and θ are the polar coordinates;  r=(Sqrt[(x^2)+(y^2)]) is the distance from the origin to the point and θ is the angle measured counterclockwise from the positive x axis to the point.  The definitions of the trig functions, for general angles, are given by

sin θ=(y/r),   cos θ=(x/r)   and   tan θ=(y/x).

Using the above definitions it is a straightforward matter to find the formulas for converting between polar and rectangular coordinates.

r and θ  ⟹  x and y

x= r cos θ   and   y=r sin θ

x and y  ⟹  r and θ

r=(Sqrt[(x^2)+(y^2)])   and   θ={({{(tan^(-1))((y/x))  when  x>0}, {180°+(tan^(-1))((y/x))  when  x<0}})

The subtlety in solving for the angle follows from the identity:  tan θ=tan(180°+θ).  The range of the (tan^(-1)) (also written arctan) function is -90°<θ<90°, which corresponds to the quadrants I and IV; this is equivalent to the condition that x>0.  The case of quadrants II and III, when x<0, requires shifting the result of the inverse tangent by 180°.

Example . - Polar Coordinates

Solution

r=(Sqrt[(x^2)+(y^2)])=(Sqrt[20])=4.47

θ=180°+(tan^(-1))((y/x))=116.6°

Vector Basics

A vector is a quantity with both a magnitude and a direction.  A scalar has only a magnitude; it is just a real number.  The magnitude of a vector is a non-negative (positive or zero) scalar.  Velocity is a vector quantity and speed is its magnitude.  Acceleration, force and momentum are also vectors.  Time, temperature, mass and pressure are examples of scalars.

We will write a vector variable by a symbol with an "⇀" over it.  The magnitude of a vector is given by the symbol without the arrow or by applying the "‖ ‖" brackets to the vector.  If  (OverVector[A]) is some vector then A=‖(OverVector[A])‖ is its magnitude.

We can represent vectors by arrows.  Suppose someone walks from a starting point (Subscript[P, 1]) to a stopping point (Subscript[P, 2]).  A displacement vector (OverVector[s]) (or Δ(OverVector[r])) may be viewed as an arrow with its tail at (Subscript[P, 1]) and its tip at (Subscript[P, 2]).

A position vector (OverVector[r]) is a displacement vector with its tail at the origin.  For more general vectors we represent them by arrows pointing in the direction of the vectors and the length of the arrow is proportional to the magnitude of the vector.  For instance, if one velocity vector has a magnitude of 60 mi/hr and another has 30 mi/hr then the arrow representing the first should have twice the length of the second.

A vector has no fixed position.  If a vector arrow is moved keeping its length and direction fixed then it still is the same vector.

Component Definition - Position and Displacement Vectors

A position vector is a way to label a position in the Cartesian plane; it has its tail at the origin and its head at the position it labels.  We will use an angled bracket notation for vectors.  The position vector that labels the point (x,y) will be written as  (OverVector[r])=⟨x,y⟩.

We will define the magnitude and direction of a vector so that the polar coordinates, r and θ,  are the magnitude and direction of the vector.

In the Cartesian plane we will denote vector from ((Subscript[x, 1]),(Subscript[y, 1])) to ((Subscript[x, 2]),(Subscript[y, 2])) by:

(OverVector[s])=⟨(Subscript[x, 2])-(Subscript[x, 1]),(Subscript[y, 2])-(Subscript[y, 1])⟩

This is called a displacement vector.  A position vector is clearly a displacement vector with its tail at the origin.

Component Definition - General Vectors

We will write a 2-dimensional vector (OverVector[A]) as a pair of real numbers (Subscript[A, x]) and (Subscript[A, y]) called components.  (A 3D vector is a triple.)  We will use the "angled-bracket" notation for vectors.

(OverVector[A])=⟨(Subscript[A, x]),(Subscript[A, y])⟩

The component (Subscript[A, x]) has the interpretation as the amount the vector (OverVector[A]) is in the x-direction and (Subscript[A, y]) the y-direction.

Magnitude and Direction Angle

We define (Subscript[A, x]) and (Subscript[A, y]) as the components of the vector (OverVector[A]).  (Subscript[A, x]) is the part of (OverVector[A]) in the x direction and similarly (Subscript[A, y]) is the y part. The Cartesian coordinates x and y are the components of a position vector (OverVector[r]). For a two dimensional vector we can represent the direction with an angle, measured as in the polar coordinates. To convert between the magnitude and direction angle and the components of a two dimensional vector we have analogous expressions to the ones for polar coordinates.

A  and  θ  ⟹  (Subscript[A, x]) and (Subscript[A, y])

(Subscript[A, x])= A cos θ   and   (Subscript[A, y])=A sin θ

(Subscript[A, x]) and (Subscript[A, y])  ⟹  A and θ

A=(Sqrt[((Subscript[A, x])^2)+((Subscript[A, y])^2)])   and   θ={({{(tan^(-1))(((Subscript[A, y])/(Subscript[A, x])))  when  (Subscript[A, x])>0}, {180°+(tan^(-1))(((Subscript[A, y])/(Subscript[A, x])))  when  (Subscript[A, x])<0}})

Vector Addition

Suppose a displacement vector (Subscript[(OverVector[s]), 1]) corresponds to someone walking from (Subscript[P, 1]) to (Subscript[P, 2]).  Suppose that then the person walks from (Subscript[P, 2]) to (Subscript[P, 3]); call this displacement (Subscript[(OverVector[s]), 2]).  The net displacement is the vector from  (Subscript[P, 1]) to (Subscript[P, 3]); this is what we will define as the sum of the two displacements (Subscript[(OverVector[s]), 1])+(Subscript[(OverVector[s]), 2]).  To generalize this to any vectors, we will define the sum of general vectors (OverVector[A]) and (OverVector[B]).  Draw the vectors as shown, with the tail of (OverVector[B]) at the tip of (OverVector[A]).  The sum  the vectors (OverVector[A])+(OverVector[B]) is the vector drawn from the tail of (OverVector[A]) to the tip of (OverVector[B]).

With our component definition vector addition takes the very simple form:

(OverVector[A])+(OverVector[B])=⟨(Subscript[A, x]),(Subscript[A, y])⟩+⟨(Subscript[B, x]),(Subscript[B, y])⟩=⟨(Subscript[A, x])+(Subscript[B, x]),(Subscript[A, y])+(Subscript[B, y])⟩.

Commutative Property

An algebraic operation is commutative when changing the order of the items doesn't affect the result.  For vector addition this takes the form.

(OverVector[A])+(OverVector[B])=(OverVector[B])+(OverVector[A])

The commutative property of the addition of reals implies this for vectors.

Because of the commutative property there are two more ways of adding vectors we can consider.  In addition to placing the tail of (OverVector[B]) at the tip of (OverVector[A]), we can place the tail of (OverVector[A]) at the tip of (OverVector[B]).  Also there is the parallelogram rule: Draw the two vector together tail to tail and complete the parallelogram; the sum it the vector from the common tail of the vectors to the opposite corner.

Associative Property

From the definition of vector addition it is clear it satisfies.

((OverVector[A])+(OverVector[B]))+(OverVector[C])=(OverVector[A])+((OverVector[B])+(OverVector[C]))

This property is called associativity.  For an associative operation it is not necessary to use brackets, since the order of the operation is unimportant.

Identity

The identity vector (OverVector[0]) is the vector that leaves any other vector (OverVector[A]) unchanged under addition.

(OverVector[A])+(OverVector[0])=(OverVector[A])

It is clear that the zero vector has zeros as components.

(OverVector[0])=⟨0,0⟩

The magnitude of the identity vector is 0, so we can write 0=‖ (OverVector[0]) ‖.  Note that the direction of (OverVector[0]) is undefined; in fact, it is the only vector with an undefined direction.

Additive Inverse.

For any vector (OverVector[A]) there is an additive inverse vector  -(OverVector[A])  with the property:

(OverVector[A])+(-(OverVector[A]))=(OverVector[0])

Clearly, this has the value

-(OverVector[A])=⟨-(Subscript[A, x]),-(Subscript[A, y])⟩

and has the same magnitude and is in the opposite direction.

Vector Subtraction

We define vector subtraction by adding the inverse.

(OverVector[A])-(OverVector[B])=(OverVector[A])+(-(OverVector[B]))

In terms of components we have

(OverVector[A])-(OverVector[B])=⟨(Subscript[A, x]),(Subscript[A, y])⟩-⟨(Subscript[B, x]),(Subscript[B, y])⟩=⟨(Subscript[A, x])-(Subscript[B, x]),(Subscript[A, y])-(Subscript[B, y])⟩.

The simplest way to view the vector (OverVector[A])-(OverVector[B]) is as the vector that when added to  (OverVector[B])  gives  (OverVector[A]); if the vectors are drawn tail to tail then it is the vector from the tip of  (OverVector[B]) to the tip of  (OverVector[A])

Scalar Multiplication

If (OverVector[A]) is a vector and c is a scalar then we can define their product c (OverVector[A]) as a vector.

c (OverVector[A])=⟨c (Subscript[A, x]),c (Subscript[A, y])⟩

It is clear that its magnitude is given by

‖c (OverVector[A])‖=|c| ‖(OverVector[A])‖,

where |c| is the absolute value of the scalar.  The direction of c (OverVector[A]) is the same as (OverVector[A]) when c>0 and opposite to (OverVector[A]) when c<0.  When c=0 we get  0 (OverVector[A])=(OverVector[0]).  Note also that 1 (OverVector[A])=(OverVector[A]) and (-1) (OverVector[A])=-(OverVector[A]).

The scalar multiplication operation has the associative and distributive properties.

Associative Property

(c d) (OverVector[A])=c (d (OverVector[A]))

Distributive Properties

(c+d) (OverVector[A])=c (OverVector[A])+d (OverVector[A])  and  c ((OverVector[A])+(OverVector[B]))=c (OverVector[A])+c (OverVector[B])

Unit Vectors and Notation

A unit vector is a vector of magnitude one.  We denote unit vectors with a " ^ " over its top.  For any vector (OverVector[A]) we can simply find the unit vector in its direction  (OverHat[A])  by

(OverHat[A])=((OverVector[A])/(‖(OverVector[A])‖)).

Basis unit vectors are unit vectors along the coordinate axes.  We will use (OverHat[x])  and (OverHat[y]) for the unit vectors in the x and y directions.  The more traditional notation for these unit vectors is to write them as  (OverHat[i]) and (OverHat[j]).

(OverHat[x])=(OverHat[i])=⟨1,0⟩  and  (OverHat[y])=(OverHat[j])=⟨0,1⟩

Any vector can then be written in terms of these basis unit vectors

(OverVector[A])=⟨(Subscript[A, x]),(Subscript[A, y])⟩=(Subscript[A, x])(OverHat[x])+(Subscript[A, y])(OverHat[y])=(Subscript[A, x])(OverHat[i])+(Subscript[A, y])(OverHat[j])

3D Vectors

It is straightforward to generalize our two dimensional discussion to a three dimensional one.  We need to add  (OverHat[z])=(OverHat[k]) for the unit vector in the z direction.

(OverHat[x])=(OverHat[i])=⟨1,0,0⟩ ,  (OverHat[y])=(OverHat[j])=⟨0,1,0⟩  and  (OverHat[z])=(OverHat[k])=⟨0,0,1⟩

We can similarly write any vector in terms or its components and unit vectors.

(OverVector[A])=⟨(Subscript[A, x]),(Subscript[A, y]),(Subscript[A, z])⟩=(Subscript[A, x])(OverHat[x])+(Subscript[A, y])(OverHat[y])+(Subscript[A, z])(OverHat[z])=(Subscript[A, x])(OverHat[i])+(Subscript[A, y])(OverHat[j])+(Subscript[A, z])(OverHat[k])

The magnitude of a vector is

A=‖(OverVector[A])‖=(Sqrt[((Subscript[A, x])^2)+((Subscript[A, y])^2)+((Subscript[A, z])^2)]).

The only subtlety with three dimensions is specifying directions as angles.  It takes two angles to specify a direction is 3D; these are the θ and ϕ of spherical coordinates.  To avoid these issues the best way to represent a direction for a three dimensional vector is to just write a unit vector in the direction of that vector

(OverHat[A])=((OverVector[A])/(‖(OverVector[A])‖)).

Example . - Vector Addition

Solution

(OverVector[A])=⟨(Subscript[A, x]),(Subscript[A, y])⟩=⟨0,-500(QuantityUnit)⟩ .

(OverVector[B])=⟨(Subscript[B, x]),(Subscript[B, y])⟩=⟨B cos (Subscript[θ, B]),B sin (Subscript[θ, B])⟩=⟨-578.51,689.44⟩(QuantityUnit) .

({{(OverVector[R])=(OverVector[A])+(OverVector[B])=⟨(Subscript[R, x]),(Subscript[R, y])⟩}, {=⟨0,-500⟩(QuantityUnit)+⟨-578.51,689.44⟩(QuantityUnit)}, {=⟨-578.51,189.44⟩(QuantityUnit)}})

R=(Sqrt[(Subscript[R, x]^2)+(Subscript[R, y]^2)])=609(QuantityUnit)                    (Magnitude)

θ=180°+(tan^(-1))((Subscript[R, y])/(Subscript[R, x]))=161.9°                 (Direction Angle)

(OverVector[R])=⟨-579,189⟩(QuantityUnit)                              (Net Displacement)

. - Kinematics in 2D and 3D

The General Problem

Position as a Function of Time

There are two equivalent ways of describing position in two dimensions.  One is by labeling the coordinates (x,y).  The other is by giving the position vector (OverVector[r]) of the position.  The two are related; the coordinates are the components of the position vector.  To label position as a function of time we can consider x and y as separate functions of time or as (OverVector[r]) as a function of time.

x(t)  and  y(t)   ⟺   (OverVector[r])(t)=⟨x(t),y(t)⟩

The actual path followed by the body is called the trajectory.  It is represented by a plot of a path in the xy plane.

Average Velocity

The average velocity, as we saw in the one dimensional case, refers to two times. At (Subscript[t, i]) the position vector is (Subscript[(OverVector[r]), i])=(OverVector[r])((Subscript[t, i])) and at (Subscript[t, f]) it is (Subscript[(OverVector[r]), f])=(OverVector[r])((Subscript[t, f])).  The displacement is the difference of these two positions

Δ (OverVector[r])=(Subscript[(OverVector[r]), f])-(Subscript[(OverVector[r]), i])=⟨(Subscript[x, f])-(Subscript[x, i]),(Subscript[y, f])-(Subscript[y, i])⟩.

The average velocity vector is then defined as

(Subscript[(OverVector[v]), ave])=((Δ (OverVector[r]))/(Δ t))=⟨((Δ x)/(Δ t)),((Δ y)/(Δ t))⟩=⟨(Subscript[v, ave,x]) ,(Subscript[v, ave,y])⟩.

Instantaneous Velocity

The instantaneous velocity is defined as the limit of the average velocity as Δ t approaches zero; it is the time derivative of the position vector.

(OverVector[v])=((d(OverVector[r]))/(dt))=(lim_(Δ t->0))((Δ(OverVector[r]))/(Δ t))=⟨((dx)/(dt)),((dy)/(dt))⟩=⟨(Subscript[v, x]),(Subscript[v, y])⟩.

The magnitude of the velocity is called the speed.

v=‖ (OverVector[v]) ‖=(Sqrt[(Subscript[v, x]^2)+(Subscript[v, y]^2)])=speed

Average Acceleration

As with the one dimensional case acceleration is to velocity as the velocity is to position.

(Subscript[(OverVector[a]), ave])=((Δ (OverVector[v]))/(Δ t))=⟨((Δ (Subscript[v, x]))/(Δ t)),((Δ (Subscript[v, y]))/(Δ t))⟩=⟨(Subscript[a, ave,x]),(Subscript[a, ave,y])⟩.

Instantaneous Acceleration

The instantaneous acceleration is similarly defined as a limit of the average acceleration or simply as the time derivative of the velocity.

(OverVector[a])=((d(OverVector[v]))/(dt))=(lim_(Δ t->0))((Δ (OverVector[v]))/(Δ t))=⟨((d(Subscript[v, x]))/(dt)),((d(Subscript[v, y]))/(dt))⟩=⟨(Subscript[a, x]),(Subscript[a, y])⟩.

Acceleration with Constant Speed

The acceleration is the time derivative of the velocity. Constant velocity implies zero acceleration. Speed is the magnitude of the acceleration so there can be an acceleration if the speed is constant, because the direction can change. If the speed is constant then the acceleration must be perpendicular to the direction of motion, which is the direction of the velocity.

constant speed    ⟺    (OverVector[a]) ⊥ (OverVector[v])

If a car is driving at a constant speed and turning left then the acceleration is toward the left.

Parallel and Perpendicular Components of Acceleration

If the speed is changing then there is a forward or backward component of the acceleration, where an increasing speed gives a forward component and a decreasing speed gives a backward component. If the speed and direction are changing then you have both components parallel and perpendicular to the direction of motion. Suppose a car is braking while turning left. There are then two components of the acceleration, the parallel component points backward and the perpendicular component points to the right.

Constant Velocity and Acceleration

The cases of constant velocity and acceleration follows the one dimensional example.  Since we are now dealing with vectors the arbitrary constants introduced in antidifferentiation become vector quantities.

const (OverVector[v])    ⟹    (OverVector[a])=(OverVector[0])  and  (OverVector[r])(t)=(Subscript[(OverVector[r]), 0])+(OverVector[v]) t

const (OverVector[a])    ⟹    (OverVector[v])(t)=(Subscript[(OverVector[v]), 0])+(OverVector[a]) t  and  (OverVector[r])(t)=(Subscript[(OverVector[r]), 0])+(Subscript[(OverVector[v]), 0]) t+(1/2)(OverVector[a]) (t^2)

For 2D motion with constant acceleration we can modify the four constant acceleration equations from Chapter B into two sets of four equations.

({{x Equations,          y  Equations}, {(Subscript[v, x])=(Subscript[v, 0,x])+(Subscript[a, x]) t, (Subscript[v, y])=(Subscript[v, 0,y])+(Subscript[a, y]) t}, {Δ x=(1/2) ((Subscript[v, x])+(Subscript[v, 0,x])) t, Δ y=(1/2) ((Subscript[v, y])+(Subscript[v, 0,y])) t}, {Δ x=(Subscript[v, 0,x]) t +(1/2) (Subscript[a, x]) (t^2), Δ y=(Subscript[v, 0,y])t +(1/2) (Subscript[a, y]) (t^2)}, {(Subscript[v, x]^2)=(Subscript[v, 0,x]^2)+2 (Subscript[a, x]) Δ x, (Subscript[v, y]^2)=(Subscript[v, 0,y]^2)+2 (Subscript[a, y]) Δ y.}})

Projectile Motion

An important case of motion with constant acceleration is that of projectile motion. Projectile motion is two dimensional motion of body under the influence of only gravity. If gravity is all that acts means that we ignore air resistance. Spin effects like a curving baseball or a slicing golf ball are associated with air resistance and will also be ignored.

For projectile motion we will take x as the horizontal direction and y as the vertical direction. The acceleration due to gravity is downward and of magnitude g.  Writing this as a vector gives

(OverVector[a])=-g (OverHat[y])=⟨0,-g⟩    or    (Subscript[a, x])=0  and  (Subscript[a, y])=-g

Inserting these components into the two sets of equations above gives:

({{Horizontal, Vertical}, {Equations, Equations}, {(Subscript[v, x])=(Subscript[v, 0,x]), (Subscript[v, y])=(Subscript[v, 0,y])-g  t}, {Δ x=(Subscript[v, 0,x]) t , Δ y=(1/2) ((Subscript[v, y])+(Subscript[v, 0,y])) t}, {, Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)}, {, (Subscript[v, y]^2)=(Subscript[v, 0,y]^2)-2 g Δ y.}})

The horizontal motion is simple.  Since there is no horizontal (the x direction) acceleration, the x component of the velocity is constant.  The vertical part (the y direction) of the motion is equivalent to free fall. The key to solving projectile motion problems is keeping the two parts separate.

If a projectile is launched at an initial angle of θ with an initial speed (Subscript[v, 0]) then the components of the initial velocity are given by

(Subscript[v, 0,x])=(Subscript[v, 0]) cos θ  and  (Subscript[v, 0,y])=(Subscript[v, 0]) sin θ.

Example . - An Initial Horizontal Velocity

Solution

Solution

h=1.35(QuantityUnit) ,  Δ x=2.42(QuantityUnit)  and  g=9.80(QuantityUnit)/(((QuantityUnit))^2)

(Subscript[v, 0])=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, 0,y]^2)])=(Sqrt[(Subscript[v, 0,x]^2)+(0^2)])=(Subscript[v, 0,x])

Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)  ⟹  -h=0-(1/2) g (t^2)  ⟹  h=(1/2) g (t^2)  ⟹  t=(Sqrt[((2 h)/g)])=0.52489(QuantityUnit)

Δ x=(Subscript[v, 0,x]) t   ⟹  (Subscript[v, 0,x])=Δ x/t=4.67(QuantityUnit)/(QuantityUnit)

Solution

v=(Sqrt[(Subscript[v, x]^2)+(Subscript[v, y]^2)])=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, y]^2)])

(Subscript[v, y])=(Subscript[v, 0,y])-g  t=-g t=-5.1438(QuantityUnit)/(QuantityUnit)

v=(Sqrt[(Subscript[v, 0,x]^2)+(Subscript[v, y]^2)])=6.95(QuantityUnit)/(QuantityUnit)

Solution

θ=(tan^(-1))((Subscript[v, y])/(Subscript[v, 0,x]))=-47.7°

Example . - Punted Football

Solution

(Subscript[v, 0])=27(QuantityUnit)/(QuantityUnit)   and   θ=63°

(Subscript[v, 0,x])=(Subscript[v, 0]) cos θ=12.258(QuantityUnit)/(QuantityUnit)  and  (Subscript[v, 0,y])=(Subscript[v, 0]) sin θ=24.057(QuantityUnit)/(QuantityUnit)

(Subscript[v, y]^2)=(Subscript[v, 0,y]^2)-2 g Δ y  ⟹  0=(Subscript[v, 0,y]^2)-2 g (Subscript[y, max])  ⟹  (Subscript[y, max])=((Subscript[v, 0,y]^2)/(2 g))=29.5(QuantityUnit)

Solution

Δ y=(Subscript[v, 0,y]) t -(1/2) g (t^2)  ⟹  0=(Subscript[v, 0,y]) t -(1/2) g (t^2)=t ((Subscript[v, 0,y])-(1/2) g t)

0=(Subscript[v, 0,y])-(1/2) g T  ⟹  T=((2 (Subscript[v, 0,y]))/g)=4.9096(QuantityUnit)=4.91(QuantityUnit)

Solution

R=Δ x=(Subscript[v, 0,x])t=(Subscript[v, 0,x])T=60.2(QuantityUnit)

To solve for the trajectory we can choose, for simplicity, that the motion begins at the origin (Subscript[x, 0])=0=(Subscript[y, 0]) giving  Δ x=x  and  Δ y=y.  Solving the horizontal equation for time gives  t=x/(Subscript[v, 0,x]).  Inserting this into the vertical expression  y=(Subscript[v, 0,y]) t -(1/2) g (t^2) gives

y=((Subscript[v, 0,y])/(Subscript[v, 0,x]))x -(g/(2 (Subscript[v, 0,x]^2))) (x^2).

It is clear that this is a parabola.

The range R of a projectile is the total horizontal distance traveled in the air when it returns to its original level, y=0 in the expression above.  We can then factor out an x from the expression and solve for x.

x=0  and  x=((2 (Subscript[v, 0,x]) (Subscript[v, 0,y]))/g)=((2 (Subscript[v, 0,])cos θ (Subscript[v, 0,]) sin θ)/g)=(((Subscript[v, 0,]^2) )/g)2 cos θ sin θ.

The x=0 solution is trivial.  Setting the other to R, writing the components in terms of (Subscript[v, 0]) and θ, and using the identity sin 2 θ=2 sin θ cos θ gives an expression for the range.

R=((Subscript[v, 0]^2)/g) sin(2 θ)

Example . - Punted Football (continued)

Solution

R=((Subscript[v, 0]^2)/g) sin 2 θ=60.2(QuantityUnit)

. - Relative Motion

A frame of reference is some coordinate system used to study motion.  Suppose there is a fixed frame K and a moving frame K',  that moves with a velocity (Subscript[(OverVector[v]), 0]) with respect to K.   If we take a moving body to be at point P.  If (OverVector[r]) is the vector from the origin of K to P,   (OverVector[r])' is the vector from the origin of K' to P,  and  (Subscript[(OverVector[r]), 0]) is the vector from the origin of K to the origin of K'.

It follows that the three position vectors are related by

(OverVector[r])=(Subscript[(OverVector[r]), 0])+(OverVector[r])'.

We want to relate the velocities of the moving body with respect to these two frames.  The velocities with respect to K and K'are (OverVector[v]) and (OverVector[v])',  respectively.  We can relate the velocities in these two frames by taking the time derivative of the expression above:

(OverVector[v])=(Subscript[(OverVector[v]), 0])+(OverVector[v])',

where (Subscript[(OverVector[v]), 0]) is the velocity of the moving frame.

Example . - Bubba

Solution

v=(Subscript[v, 0])+v'=85(QuantityUnit)/(QuantityUnit)+15(QuantityUnit)/(QuantityUnit)=100(QuantityUnit)/(QuantityUnit)

Solution

v=(Subscript[v, 0])+v'=85(QuantityUnit)/(QuantityUnit)+(-15(QuantityUnit)/(QuantityUnit))=70(QuantityUnit)/(QuantityUnit)

Solution

v=(Sqrt[(((85(QuantityUnit)/(QuantityUnit)))^2)+(((15(QuantityUnit)/(QuantityUnit)))^2)])=86.3(QuantityUnit)/(QuantityUnit)

Solution

θ=arctan ((15(QuantityUnit)/(QuantityUnit))/(85(QuantityUnit)/(QuantityUnit)))=10.0°

Example . - A Plane in a Cross-wind

Solution

(Subscript[(OverVector[v]), 0])=⟨0,-100⟩(QuantityUnit)/(QuantityUnit)

θ=(sin^(-1))(100/400)=14.5°

Solution

(v^2)+(Subscript[v, 0]^2)=((v')^2)  ⟹  v=(Sqrt[((v')^2)-(Subscript[v, 0]^2)])=(Sqrt[(((400(QuantityUnit)/(QuantityUnit)))^2)-(((100(QuantityUnit)/(QuantityUnit)))^2)])=387.30(QuantityUnit)/(QuantityUnit)

d=v t   ⟹  t=(d/v)=2.07(QuantityUnit)