The Central Idea
Suppose two objects, \(A\) and \(B\), move through the same coordinate system.
Their positions are
\[ \mathbf{r}_A(t) \]and
\[ \mathbf{r}_B(t). \]The position of \(A\) relative to \(B\) is
\[ \boxed{ \mathbf{r}_{A/B} = \mathbf{r}_A-\mathbf{r}_B }. \]Relative motion is built from this difference.
What Does “Relative To” Mean?
The phrase relative to identifies the observer or reference object.
The notation
\[ \mathbf{r}_{A/B} \]means:
“the position of \(A\) as measured relative to \(B\).”
The order matters.
Relative Position Reverses Sign When the Order Reverses
Since
\[ \mathbf{r}_{A/B} = \mathbf{r}_A-\mathbf{r}_B, \]reversing the viewpoint gives
\[ \mathbf{r}_{B/A} = \mathbf{r}_B-\mathbf{r}_A. \]Therefore,
\[ \boxed{ \mathbf{r}_{B/A} = -\mathbf{r}_{A/B} }. \]Relative Position in One Dimension
On a number line, suppose
\[ x_A=12 \]and
\[ x_B=5. \]Then the position of \(A\) relative to \(B\) is
\[ x_{A/B}=12-5=7. \]Thus \(A\) is 7 units in the positive direction from \(B\).
Example: Reversing the Viewpoint
Using the same positions,
\[ x_A=12, \qquad x_B=5, \]the position of \(B\) relative to \(A\) is
\[ x_{B/A}=5-12=-7. \]Therefore,
\[ \boxed{ x_{A/B}=7, \qquad x_{B/A}=-7 }. \]Separation Distance
Relative position is directional.
Separation distance is its magnitude.
In one dimension,
\[ \boxed{ d_{AB} = |x_A-x_B| }. \]In vector form,
\[ \boxed{ d_{AB} = |\mathbf{r}_A-\mathbf{r}_B| }. \]Relative Position in Two Dimensions
Suppose
\[ \mathbf{r}_A = \langle7,4\rangle \]and
\[ \mathbf{r}_B = \langle2,-1\rangle. \]Then
\[ \mathbf{r}_{A/B} = \langle7,4\rangle-\langle2,-1\rangle \] \[ = \boxed{\langle5,5\rangle}. \]Distance between Two Moving Objects
If
\[ \mathbf{r}_{A/B} = \langle5,5\rangle, \]then the separation distance is
\[ |\mathbf{r}_{A/B}| = \sqrt{5^2+5^2} \] \[ = \boxed{5\sqrt2}. \]Relative Velocity
Relative velocity is the rate of change of relative position.
Differentiate:
\[ \mathbf{r}_{A/B} = \mathbf{r}_A-\mathbf{r}_B. \]Therefore,
\[ \boxed{ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B }. \]Relative Velocity Reverses Sign with Viewpoint
Since
\[ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B, \]reversing the order gives
\[ \boxed{ \mathbf{v}_{B/A} = -\mathbf{v}_{A/B} }. \]Example: Same-Direction Motion
Two cars move east along the same road.
Car \(A\) travels at
\[ 70\frac{\text{km}}{\text{h}}, \]while car \(B\) travels at
\[ 50\frac{\text{km}}{\text{h}}. \]Taking east as positive,
\[ v_{A/B}=70-50. \]Therefore,
\[ \boxed{ v_{A/B} = 20\frac{\text{km}}{\text{h}} }. \]To an observer moving with \(B\), car \(A\) appears to move east at \(20\text{ km/h}\).
Closing Speed in the Same Direction
If a faster object is behind a slower object moving in the same direction, the gap closes at the difference of their speeds.
If
\[ v_{\text{fast}}>v_{\text{slow}}, \]then
\[ \boxed{ \text{closing speed} = v_{\text{fast}}-v_{\text{slow}} }. \]Example: Catch-Up Time
A faster car travels at
\[ 80\frac{\text{km}}{\text{h}} \]behind a slower car traveling at
\[ 60\frac{\text{km}}{\text{h}}. \]The slower car has a 30-kilometer lead.
Closing speed:
\[ 80-60=20\frac{\text{km}}{\text{h}}. \]Catch-up time:
\[ t=\frac{30}{20}. \]Therefore,
\[ \boxed{t=1.5\text{ h}}. \]Opposite-Direction Motion
Suppose east is positive.
If object \(A\) moves east at
\[ 40\frac{\text{km}}{\text{h}} \]and object \(B\) moves west at
\[ 30\frac{\text{km}}{\text{h}}, \]then
\[ v_A=40, \qquad v_B=-30. \]Therefore,
\[ v_{A/B} = 40-(-30) \] \[ = \boxed{ 70\frac{\text{km}}{\text{h}} }. \]Closing Speed for Objects Moving toward Each Other
When two objects move directly toward each other along the same line, the separation decreases at the sum of their speeds.
Thus
\[ \boxed{ \text{closing speed} = s_1+s_2 }. \]This rule follows from signed velocity subtraction.
Example: Meeting Time
Two vehicles are 210 kilometers apart and travel directly toward each other.
Their speeds are
\[ 50\frac{\text{km}}{\text{h}} \]and
\[ 55\frac{\text{km}}{\text{h}}. \]Their closing speed is
\[ 50+55=105\frac{\text{km}}{\text{h}}. \]Therefore,
\[ t=\frac{210}{105} \] \[ \boxed{t=2\text{ h}}. \]Objects Moving Apart
If two objects move away from one another, the separation distance increases.
Along a straight line, the separation rate can often be determined from the magnitude of the appropriate relative velocity.
Direction and sign should still be handled consistently before taking magnitudes.
Relative Motion from Position Functions
Suppose
\[ x_A(t)=4t+2 \]and
\[ x_B(t)=t+8. \]Then the relative position of \(A\) with respect to \(B\) is
\[ x_{A/B}(t) = (4t+2)-(t+8) \] \[ = \boxed{3t-6}. \]When Do Two Objects Meet?
Two objects meet when their positions are equal:
\[ \mathbf{r}_A(t)=\mathbf{r}_B(t). \]Equivalently,
\[ \boxed{ \mathbf{r}_{A/B}(t)=\mathbf{0} }. \]In one dimension,
\[ \boxed{ x_A(t)-x_B(t)=0 }. \]Example: Solving for a Meeting Time
Continue with
\[ x_{A/B}(t)=3t-6. \]The objects meet when
\[ 3t-6=0. \]Therefore,
\[ \boxed{t=2}. \]Relative Velocity from Position Functions
Differentiate the relative position:
\[ x_{A/B}(t)=3t-6. \]Then
\[ v_{A/B}(t)=3. \]This also agrees with
\[ v_A-v_B=4-1=3. \]Relative Acceleration
Relative acceleration is the rate of change of relative velocity.
Therefore,
\[ \boxed{ \mathbf{a}_{A/B} = \mathbf{a}_A-\mathbf{a}_B }. \]Like relative position and velocity, it is found by subtraction.
Example: Relative Acceleration
Suppose
\[ \mathbf{a}_A = \langle3,1\rangle \]and
\[ \mathbf{a}_B = \langle1,-2\rangle. \]Then
\[ \mathbf{a}_{A/B} = \langle3,1\rangle-\langle1,-2\rangle \] \[ = \boxed{\langle2,3\rangle}. \]The Relative-Motion Chain
Relative motion follows the same derivative structure as ordinary motion:
\[ \mathbf{r}_{A/B} \xrightarrow{\text{differentiate}} \mathbf{v}_{A/B} \xrightarrow{\text{differentiate}} \mathbf{a}_{A/B}. \]Symbolically,
\[ \boxed{ \mathbf{v}_{A/B} = \frac{d}{dt}\mathbf{r}_{A/B} } \]and
\[ \boxed{ \mathbf{a}_{A/B} = \frac{d}{dt}\mathbf{v}_{A/B} }. \]Moving Observers
Relative velocity describes what one moving observer sees another object doing.
If observer \(B\) moves with velocity \(\mathbf{v}_B\), then object \(A\) appears to move with velocity
\[ \boxed{ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B }. \]This difference removes the motion shared by both objects.
Example: Two Trains
Two trains move east on parallel tracks.
Train \(A\) moves at
\[ 90\frac{\text{km}}{\text{h}}, \]while train \(B\) moves at
\[ 85\frac{\text{km}}{\text{h}}. \]To a passenger on train \(B\), train \(A\) appears to move east at
\[ 90-85 \] \[ = \boxed{ 5\frac{\text{km}}{\text{h}} }. \]Example: Apparent Motion in Opposite Directions
Train \(A\) moves east at
\[ 80\frac{\text{km}}{\text{h}}, \]while train \(B\) moves west at
\[ 70\frac{\text{km}}{\text{h}}. \]Taking east as positive,
\[ v_A=80, \qquad v_B=-70. \]Therefore,
\[ v_{A/B} = 80-(-70) \] \[ = \boxed{ 150\frac{\text{km}}{\text{h}} }. \]Relative Velocity in Two Dimensions
Suppose
\[ \mathbf{v}_A = \langle6,2\rangle \]and
\[ \mathbf{v}_B = \langle1,5\rangle. \]Then
\[ \mathbf{v}_{A/B} = \langle6,2\rangle-\langle1,5\rangle \] \[ = \boxed{\langle5,-3\rangle}. \]Magnitude of Relative Velocity
If
\[ \mathbf{v}_{A/B} = \langle5,-3\rangle, \]then its magnitude is
\[ |\mathbf{v}_{A/B}| = \sqrt{5^2+(-3)^2} \] \[ = \boxed{\sqrt{34}}. \]This gives the magnitude of \(A\)'s velocity as observed from \(B\)'s translating frame.
Relative Motion and Wind
Relative velocity is useful when one motion occurs within another moving medium.
For example, an aircraft's velocity relative to the ground can be modeled as
\[ \boxed{ \mathbf{v}_{\text{plane/ground}} = \mathbf{v}_{\text{plane/air}} + \mathbf{v}_{\text{air/ground}} }. \]This is a vector-addition version of relative motion.
Why the Wind Formula Uses Addition
Relative-velocity notation obeys a chain rule:
\[ \mathbf{v}_{A/C} = \mathbf{v}_{A/B} + \mathbf{v}_{B/C}. \]For a plane, air, and ground,
\[ \boxed{ \mathbf{v}_{\text{plane/ground}} = \mathbf{v}_{\text{plane/air}} + \mathbf{v}_{\text{air/ground}} }. \]The intermediate reference \(B\) cancels conceptually.
Relative-Velocity Chain Rule
In general,
\[ \boxed{ \mathbf{v}_{A/C} = \mathbf{v}_{A/B} + \mathbf{v}_{B/C} }. \]Since
\[ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B \]and
\[ \mathbf{v}_{B/C} = \mathbf{v}_B-\mathbf{v}_C, \]adding gives
\[ \mathbf{v}_A-\mathbf{v}_C = \mathbf{v}_{A/C}. \]Boat and Current Example
Suppose a boat moves through water at
\[ \mathbf{v}_{\text{boat/water}} = \langle4,0\rangle \]while the current moves at
\[ \mathbf{v}_{\text{water/ground}} = \langle0,3\rangle. \]Then
\[ \mathbf{v}_{\text{boat/ground}} = \langle4,0\rangle+\langle0,3\rangle \] \[ = \boxed{\langle4,3\rangle}. \]Ground Speed of the Boat
For
\[ \mathbf{v}_{\text{boat/ground}} = \langle4,3\rangle, \]the ground speed is
\[ \sqrt{4^2+3^2} \] \[ = \boxed{5}. \]Crosswind Example
An aircraft points due north with airspeed
\[ 200\frac{\text{km}}{\text{h}}. \]Let east be the positive \(x\)-direction and north the positive \(y\)-direction.
Then
\[ \mathbf{v}_{\text{plane/air}} = \langle0,200\rangle. \]If the wind blows east at
\[ 50\frac{\text{km}}{\text{h}}, \]then
\[ \mathbf{v}_{\text{air/ground}} = \langle50,0\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}_{\text{plane/ground}} = \langle50,200\rangle }. \]Ground-Speed Magnitude with Crosswind
The magnitude is
\[ \sqrt{50^2+200^2} \] \[ = \sqrt{42500} \] \[ = \boxed{50\sqrt{17}}. \]The aircraft's ground track is not due north because the wind adds an eastward component.
Pursuit Problems
In a pursuit problem, one object attempts to reach another moving object.
Along a line, the key quantities are:
- initial separation;
- pursuer velocity;
- target velocity;
- relative velocity;
- time until relative position becomes zero.
General One-Dimensional Pursuit Model
Suppose
\[ x_A(t)=x_{A0}+v_At \]and
\[ x_B(t)=x_{B0}+v_Bt. \]Then
\[ x_{A/B}(t) = (x_{A0}-x_{B0}) + (v_A-v_B)t. \]Thus
\[ \boxed{ x_{A/B}(t) = x_{A/B}(0) + v_{A/B}t }. \]Solving the Pursuit Model
Interception occurs when
\[ x_{A/B}(t)=0. \]Therefore,
\[ 0 = x_{A/B}(0) + v_{A/B}t. \]Solving,
\[ \boxed{ t = -\frac{x_{A/B}(0)}{v_{A/B}} }. \]The sign and physical meaning depend on the chosen ordering and coordinate convention.
Example: Pursuit from Behind
A target begins 100 meters ahead of a pursuer.
The target moves at
\[ 6\frac{\text{m}}{\text{s}}, \]while the pursuer moves at
\[ 11\frac{\text{m}}{\text{s}}. \]The closing speed is
\[ 11-6=5\frac{\text{m}}{\text{s}}. \]Therefore,
\[ t=\frac{100}{5} \] \[ \boxed{t=20\text{ s}}. \]Interception in Two Dimensions
In two dimensions, interception requires equality of both coordinate components at the same time.
Thus
\[ \boxed{ \mathbf{r}_A(t) = \mathbf{r}_B(t) }. \]Equivalently,
\[ \boxed{ \mathbf{r}_{A/B}(t) = \mathbf{0} }. \]Example: Two-Dimensional Interception
Let
\[ \mathbf{r}_A(t) = \langle2t,t\rangle \]and
\[ \mathbf{r}_B(t) = \langle6-t,3\rangle. \]Set the position vectors equal:
\[ 2t=6-t \]and
\[ t=3. \]The first equation also gives
\[ 3t=6, \]so
\[ t=2. \]The two component conditions disagree.
Therefore, the objects do not occupy the same point at the same time.
Closest Approach
Two moving objects need not collide.
We may instead ask when their separation is smallest.
Define the separation vector
\[ \mathbf{s}(t) = \mathbf{r}_A(t)-\mathbf{r}_B(t). \]Then the squared separation is
\[ \boxed{ D^2(t) = \mathbf{s}(t)\cdot\mathbf{s}(t) }. \]Minimizing \(D^2(t)\) is often easier than minimizing \(D(t)\).
Why Squared Distance Is Useful
Since
\[ D(t)=|\mathbf{s}(t)|, \]we have
\[ D^2(t)=|\mathbf{s}(t)|^2. \]The square root is removed, which usually simplifies differentiation and algebra.
The time minimizing \(D(t)\) also minimizes \(D^2(t)\) because squaring is increasing for nonnegative values.
Example: Closest Approach on Parallel Paths
Suppose
\[ \mathbf{r}_A(t) = \langle t,0\rangle \]and
\[ \mathbf{r}_B(t) = \langle5-t,3\rangle. \]Then
\[ \mathbf{s}(t) = \langle2t-5,-3\rangle. \]Squared separation:
\[ D^2(t) = (2t-5)^2+9. \]This is minimized when
\[ 2t-5=0. \]Therefore,
\[ \boxed{t=\frac52}. \]The minimum separation is then
\[ \boxed{3}. \]Closing Rate versus Relative Speed
In one-dimensional head-on motion, the magnitude of relative velocity is the closing speed.
In multiple dimensions, however, the magnitude
\[ |\mathbf{v}_{A/B}| \]is not always the instantaneous rate at which the separation distance changes.
Relative velocity may have both radial and sideways components.
Rate of Change of Separation Distance
Let
\[ \mathbf{s}(t) = \mathbf{r}_{A/B}(t). \]The separation distance is
\[ D(t)=|\mathbf{s}(t)|. \]When \(D(t)\ne0\),
\[ \boxed{ \frac{dD}{dt} = \frac{ \mathbf{s}\cdot\mathbf{v}_{A/B} }{ |\mathbf{s}| } }. \]This measures the component of relative velocity along the line of separation.
Interpreting the Sign of Separation Rate
If
\[ \frac{dD}{dt}<0, \]the objects are getting closer.
If
\[ \frac{dD}{dt}>0, \]they are moving farther apart.
If
\[ \frac{dD}{dt}=0, \]their separation is momentarily neither increasing nor decreasing.
Relative Motion in a Translating Frame
Suppose a coordinate frame moves with origin
\[ \mathbf{R}(t) \]relative to a fixed frame.
If an object has fixed-frame position
\[ \mathbf{r}(t), \]then its position relative to the moving origin is
\[ \boxed{ \mathbf{r}_{\text{rel}}(t) = \mathbf{r}(t)-\mathbf{R}(t) }. \]Velocity in a Translating Frame
Differentiate:
\[ \boxed{ \mathbf{v}_{\text{rel}}(t) = \mathbf{v}(t)-\mathbf{V}(t) }. \]Here \(\mathbf{V}(t)=\mathbf{R}'(t)\) is the velocity of the moving frame's origin.
This simple subtraction applies to translating frames whose axes remain parallel to the original axes.
Acceleration in a Translating Frame
Differentiate again:
\[ \boxed{ \mathbf{a}_{\text{rel}}(t) = \mathbf{a}(t)-\mathbf{A}(t) }. \]If the moving frame has constant velocity, then
\[ \mathbf{A}(t)=\mathbf{0}, \]so both frames measure the same acceleration.
Galilean Velocity Shift
If a frame moves with constant velocity \(\mathbf{V}\), then
\[ \boxed{ \mathbf{v}_{\text{moving frame}} = \mathbf{v}_{\text{fixed frame}} - \mathbf{V} }. \]This is the classical Galilean velocity transformation for nonrotating frames in ordinary low-speed mechanics.
Shared Motion Cancels
Suppose two objects both move east at
\[ 100\frac{\text{km}}{\text{h}}, \]with constant separation.
Then
\[ v_{A/B}=100-100=0. \]Each object is moving rapidly relative to the ground, but neither moves relative to the other.
Common Error: Adding Speeds in Every Relative-Motion Problem
Speeds add directly only in certain one-dimensional configurations, such as two objects moving toward each other.
The general rule is
\[ \boxed{ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B }. \]Correct signs or vector components should be used before taking magnitudes.
Common Error: Ignoring the Order of Relative Notation
The expressions
\[ \mathbf{v}_{A/B} \]and
\[ \mathbf{v}_{B/A} \]are opposites:
\[ \boxed{ \mathbf{v}_{B/A} = -\mathbf{v}_{A/B} }. \]Common Error: Confusing Relative Velocity with Separation Distance
Relative velocity describes how the relative position vector changes.
Separation distance is the magnitude of relative position:
\[ |\mathbf{r}_{A/B}|. \]These are different quantities with different units.
Common Error: Using Relative Speed as the Separation Rate in Two Dimensions
In multiple dimensions,
\[ |\mathbf{v}_{A/B}| \]measures the magnitude of relative velocity.
The actual rate of change of separation is only the component along the separation direction.
Therefore, sideways relative motion can contribute to relative speed without immediately changing the separation distance.
Common Error: Solving Only One Coordinate in an Interception Problem
In two or three dimensions, an interception requires every coordinate to match at the same time.
Solving only
\[ x_A(t)=x_B(t) \]is not enough unless the remaining coordinates also agree.
Common Error: Forgetting the Initial Separation
A pursuer may be faster than a target without catching it immediately.
Catch-up time depends on both:
- the initial gap;
- the closing speed.
The basic one-dimensional relationship is
\[ \boxed{ t = \frac{\text{initial separation}} {\text{closing speed}} } \]when the closing speed is constant and positive.
Common Error: Mixing Reference Frames
All vectors in a relative-motion equation must be interpreted consistently.
For example,
\[ \mathbf{v}_{\text{plane/air}} + \mathbf{v}_{\text{air/ground}} \]can be added because the intermediate reference frame links them correctly.
Mixing unrelated reference labels can produce meaningless results.
A Reliable Procedure for One-Dimensional Relative Motion
- Choose a positive direction.
- Assign signed positions and velocities.
- Identify which object is being measured relative to which.
- Subtract positions to form relative position.
- Subtract velocities to form relative velocity.
- Set relative position equal to zero for a meeting or interception.
- Use the sign of relative velocity to interpret closing or separation.
- Check units and physical plausibility.
A Reliable Procedure for Multi-Dimensional Relative Motion
- Write each object's position vector.
- Subtract to form \(\mathbf{r}_{A/B}\).
- Differentiate or subtract velocities to form \(\mathbf{v}_{A/B}\).
- Use \(|\mathbf{r}_{A/B}|\) for separation distance.
- Set \(\mathbf{r}_{A/B}=\mathbf{0}\) for interception.
- Check every component simultaneously.
- Minimize \(|\mathbf{r}_{A/B}|^2\) when closest approach is needed.
A Reliable Procedure for Moving-Medium Problems
- Identify the object, medium, and ground or fixed reference.
- Write each velocity with explicit relative labels.
- Arrange the labels so intermediate references connect.
- Add the vectors component by component.
- Find magnitude and direction only after obtaining the resultant vector.
- Check that the direction matches the physical situation.
Example: Full One-Dimensional Workflow
Car \(A\) begins at
\[ x_A(0)=0 \]and moves east at
\[ 25\frac{\text{m}}{\text{s}}. \]Car \(B\) begins 120 meters ahead:
\[ x_B(0)=120, \]and moves east at
\[ 19\frac{\text{m}}{\text{s}}. \]Their position functions are
\[ x_A(t)=25t \]and
\[ x_B(t)=120+19t. \]Relative position:
\[ x_{A/B}(t) = 25t-(120+19t) \] \[ = 6t-120. \]Interception occurs when
\[ 6t-120=0. \]Therefore,
\[ \boxed{t=20\text{ s}}. \]Example: Full Vector Workflow
Let
\[ \mathbf{r}_A(t) = \langle2t,3t\rangle \]and
\[ \mathbf{r}_B(t) = \langle10-t,2+t\rangle. \]Relative position:
\[ \mathbf{r}_{A/B}(t) = \langle2t-(10-t),3t-(2+t)\rangle \] \[ = \boxed{ \langle3t-10,2t-2\rangle }. \]Relative velocity:
\[ \boxed{ \mathbf{v}_{A/B} = \langle3,2\rangle }. \]For interception, both components of relative position would need to be zero at the same time.
Practice 1: Relative Position
In one dimension, \(A\) is at \(x=14\) and \(B\) is at \(x=6\). Find \(x_{A/B}\).
Practice 2: Relative Velocity in the Same Direction
Two cars move east at \(75\text{ km/h}\) and \(60\text{ km/h}\). Find the faster car's velocity relative to the slower car.
Practice 3: Opposite Directions
One vehicle moves east at \(45\text{ km/h}\) and another moves west at \(35\text{ km/h}\). Find their relative speed.
Practice 4: Catch-Up Time
A runner moving at \(8\text{ m/s}\) is 60 meters behind a runner moving at \(5\text{ m/s}\). How long does it take the faster runner to catch up?
Practice 5: Meeting Time
Two vehicles are 180 kilometers apart and travel toward each other at \(40\text{ km/h}\) and \(50\text{ km/h}\). When do they meet?
Practice 6: Relative Velocity Vector
Let
\[ \mathbf{v}_A=\langle7,4\rangle, \qquad \mathbf{v}_B=\langle2,-1\rangle. \]Find \(\mathbf{v}_{A/B}\).
Practice 7: Boat and Current
A boat moves relative to the water with
\[ \mathbf{v}_{\text{boat/water}} = \langle5,0\rangle \]while the current has
\[ \mathbf{v}_{\text{water/ground}} = \langle0,2\rangle. \]Find the boat's velocity relative to the ground.
Practice 8: Relative Position Function
Let
\[ x_A(t)=5t+1 \]and
\[ x_B(t)=2t+10. \]Find \(x_{A/B}(t)\) and the time when the objects meet.
Practice 9: Closest Approach
Let
\[ \mathbf{r}_A(t) = \langle t,0\rangle \]and
\[ \mathbf{r}_B(t) = \langle4-t,2\rangle. \]Find the time of closest approach and the minimum separation.
Practice 10: Identify the Error
A student says that two objects moving east at \(70\text{ km/h}\) and \(50\text{ km/h}\) have relative speed \(120\text{ km/h}\).
Explain the error.
Practice Answers
Practice 1
\[ x_{A/B}=14-6. \] \[ \boxed{x_{A/B}=8}. \]Practice 2
\[ 75-60 = \boxed{ 15\frac{\text{km}}{\text{h}} }. \]Practice 3
Taking east as positive,
\[ v_A=45, \qquad v_B=-35. \]Therefore,
\[ v_{A/B}=45-(-35)=80. \]The relative speed is
\[ \boxed{ 80\frac{\text{km}}{\text{h}} }. \]Practice 4
Closing speed:
\[ 8-5=3\frac{\text{m}}{\text{s}}. \]Catch-up time:
\[ t=\frac{60}{3} \] \[ \boxed{t=20\text{ s}}. \]Practice 5
Closing speed:
\[ 40+50=90\frac{\text{km}}{\text{h}}. \]Therefore,
\[ t=\frac{180}{90} \] \[ \boxed{t=2\text{ h}}. \]Practice 6
\[ \mathbf{v}_{A/B} = \langle7,4\rangle-\langle2,-1\rangle \] \[ = \boxed{\langle5,5\rangle}. \]Practice 7
\[ \mathbf{v}_{\text{boat/ground}} = \mathbf{v}_{\text{boat/water}} + \mathbf{v}_{\text{water/ground}}. \] \[ = \langle5,0\rangle+\langle0,2\rangle. \] \[ \boxed{ \mathbf{v}_{\text{boat/ground}} = \langle5,2\rangle }. \]Practice 8
\[ x_{A/B}(t) = (5t+1)-(2t+10) \] \[ = \boxed{3t-9}. \]They meet when
\[ 3t-9=0. \]Therefore,
\[ \boxed{t=3}. \]Practice 9
\[ \mathbf{r}_{A/B}(t) = \langle t-(4-t),0-2\rangle \] \[ = \langle2t-4,-2\rangle. \]Squared separation:
\[ D^2(t) = (2t-4)^2+4. \]This is minimized when
\[ 2t-4=0, \]so
\[ \boxed{t=2}. \]Then
\[ D_{\min} = \sqrt{4} \] \[ \boxed{D_{\min}=2}. \]Practice 10
The two objects move in the same direction, so their relative velocity is found by subtraction:
\[ 70-50 = \boxed{ 20\frac{\text{km}}{\text{h}} }. \]Adding the speeds would be appropriate for direct opposite-direction motion, not this same-direction case.
What to Remember
Relative position is
\[ \boxed{ \mathbf{r}_{A/B} = \mathbf{r}_A-\mathbf{r}_B }. \]Relative velocity is
\[ \boxed{ \mathbf{v}_{A/B} = \mathbf{v}_A-\mathbf{v}_B }. \]Relative acceleration is
\[ \boxed{ \mathbf{a}_{A/B} = \mathbf{a}_A-\mathbf{a}_B }. \]Separation distance is
\[ \boxed{ |\mathbf{r}_{A/B}| }. \]The most important ideas are:
- relative quantities depend on the chosen reference object;
- reversing the reference order reverses the relative vector;
- same-direction closing speed is usually a difference;
- opposite-direction head-on closing speed is usually a sum;
- interception requires relative position to become zero;
- in multiple dimensions, every coordinate condition must hold at the same time;
- moving-medium problems use linked relative-velocity vectors.
Where We Go Next
Relative motion compares moving objects directly. The next step is to examine situations where paths oppose one another, cross, or intersect, creating questions about meeting points, collision timing, and whether two trajectories reach the same location at the same time.
The next lesson, Opposing and Intersecting Paths, develops these geometric and temporal conditions systematically.