The Central Idea
Ordinary distance-rate-time problems use
\[ d=rt. \]Relative-motion problems ask us to apply that relationship to the distance between two moving objects.
If two objects move directly toward one another at speeds \(r_1\) and \(r_2\), the distance between them decreases at
\[ \boxed{r_1+r_2}. \]If one object follows another in the same direction, the gap changes at
\[ \boxed{|r_1-r_2|}. \]These are relative rates.
What Is Relative Speed?
Relative speed is the rate at which the distance between two objects changes.
It is not necessarily the speed of either object individually.
Two cars moving toward one another at \(50\) mph and \(60\) mph close the distance between them at
\[ 50+60=110. \]Therefore, their relative speed is
\[ \boxed{110\text{ mph}}. \]Opposite Directions: Add the Speeds
Suppose two objects begin at the same point and move in opposite directions at rates \(r_1\) and \(r_2\).
After time \(t\), their individual distances are
\[ d_1=r_1t \]and
\[ d_2=r_2t. \]Their separation is
\[ d_1+d_2. \]Therefore,
\[ D=r_1t+r_2t. \]Factor:
\[ \boxed{D=(r_1+r_2)t}. \]Example: Moving Apart from the Same Point
Two cyclists leave the same location at the same time in opposite directions. One travels at \(14\) mph and the other at \(18\) mph.
Their separation rate is
\[ 14+18=32\text{ mph}. \]After \(3\) hours, their separation is
\[ 32(3)=96. \]Therefore,
\[ \boxed{96\text{ miles}}. \]Meeting from Opposite Directions
Suppose two objects begin \(D\) units apart and move directly toward one another at rates \(r_1\) and \(r_2\).
If they meet after time \(t\), their traveled distances add to the original separation:
\[ r_1t+r_2t=D. \]Factor:
\[ (r_1+r_2)t=D. \]Therefore,
\[ \boxed{t=\frac{D}{r_1+r_2}}. \]Example: Meeting Time
Two cities are \(330\) miles apart. Two cars leave at the same time and drive toward one another at \(55\) mph and \(65\) mph.
Their closing speed is
\[ 55+65=120\text{ mph}. \]Therefore,
\[ t=\frac{330}{120}=2.75. \]Thus they meet after
\[ \boxed{2.75\text{ hr}=2\text{ hr }45\text{ min}}. \]Finding the Meeting Location
Continue the previous example.
The first car travels
\[ 55(2.75)=151.25 \]miles.
The second car travels
\[ 65(2.75)=178.75 \]miles.
Check:
\[ 151.25+178.75=330. \]Therefore, the meeting point is
\[ \boxed{151.25\text{ miles from the first city}}. \]Same Direction: Subtract the Speeds
When two objects move in the same direction, the faster object changes the gap relative to the slower object at the difference of their speeds.
If the faster rate is \(r_f\) and the slower rate is \(r_s\), then
\[ \boxed{r_{\text{relative}}=r_f-r_s}. \]This difference is the rate at which the faster object gains on the slower one.
Example: Same-Direction Separation
Two vehicles travel in the same direction at \(68\) mph and \(52\) mph.
The faster vehicle gains at
\[ 68-52=16\text{ mph}. \]If they begin side by side, after \(2.5\) hours their separation is
\[ 16(2.5)=40. \]Therefore,
\[ \boxed{40\text{ miles}}. \]Pursuit Problems
In a pursuit problem, one object begins ahead of another, and the faster object attempts to catch it.
The key quantity is the initial lead.
If the lead is \(L\) and the faster object gains at
\[ r_f-r_s, \]then the catch-up time is
\[ \boxed{ t=\frac{L}{r_f-r_s} }. \]Example: Catching a Slower Vehicle
A truck is \(30\) miles ahead of a car. The truck travels at \(50\) mph, and the car travels at \(65\) mph in the same direction.
The car gains at
\[ 65-50=15\text{ mph}. \]Therefore,
\[ t=\frac{30}{15}=2. \]The car catches the truck after
\[ \boxed{2\text{ hours}}. \]Why the Difference of Speeds Works
Suppose the slower object begins \(L\) units ahead.
After time \(t\), the slower object has traveled an additional
\[ r_st, \]so its position relative to the pursuer's starting point is
\[ L+r_st. \]The faster object's position is
\[ r_ft. \]At the catch point,
\[ r_ft=L+r_st. \]Rearranging:
\[ (r_f-r_s)t=L. \]Therefore,
\[ \boxed{t=\frac{L}{r_f-r_s}}. \]Head Starts Create Different Travel Times
A pursuit problem may describe a head start in time rather than distance.
Suppose a slower traveler leaves \(h\) hours before the faster traveler.
If the faster traveler moves for \(t\) hours, then the slower traveler has been moving for
\[ t+h. \]At the catch point, the traveled distances are equal:
\[ \boxed{r_f t=r_s(t+h)}. \]Example: Pursuit after a Head Start
A bus leaves a station traveling at \(45\) mph. One hour later, a car leaves the same station in the same direction at \(60\) mph.
Let \(t\) be the number of hours the car travels before catching the bus.
The car travels
\[ 60t. \]The bus travels for \(t+1\) hours:
\[ 45(t+1). \]At the catch point,
\[ 60t=45(t+1). \]Solve:
\[ 60t=45t+45, \] \[ 15t=45, \] \[ t=3. \]Therefore, the car catches the bus
\[ \boxed{3\text{ hours after the car leaves}}. \]The Same Problem Using the Initial Lead
During the bus's one-hour head start, it travels
\[ 45(1)=45 \]miles.
The car gains at
\[ 60-45=15\text{ mph}. \]Therefore,
\[ t=\frac{45}{15}=3. \]This gives the same result.
Both approaches are valid: equal-distance equations and relative-speed equations describe the same situation.
Distance Tables for Relative Motion
A table helps keep each object's rate, time, and distance separate.
\[ \begin{array}{c|c|c|c} \text{object} & \text{rate} & \text{time} & \text{distance}\\ \hline A & r_A & t_A & r_At_A\\ B & r_B & t_B & r_Bt_B \end{array} \]The equation depends on how the distances relate.
- Moving toward each other: distances add to the starting separation.
- Moving apart: distances add to the final separation.
- Pursuit: positions become equal at the catch point.
Opposing Motion with an Unknown Rate
Two vehicles begin \(280\) miles apart and move toward one another. One travels at \(50\) mph and the other at \(r\) mph. They meet in \(2.5\) hours.
Write
\[ 50(2.5)+r(2.5)=280. \]Then
\[ 125+2.5r=280, \] \[ 2.5r=155, \] \[ r=62. \]Therefore,
\[ \boxed{r=62\text{ mph}}. \]Opposing Motion with Different Start Times
Two objects may move toward one another but begin at different times.
Suppose object A begins \(h\) hours before object B.
If B travels for \(t\) hours before the meeting, then A travels for
\[ t+h. \]If the initial separation is \(D\), then
\[ \boxed{ r_A(t+h)+r_Bt=D }. \]Example: Different Departure Times
Two towns are \(300\) miles apart. A car leaves the first town at \(50\) mph. One hour later, another car leaves the second town toward the first at \(75\) mph.
Let \(t\) be the number of hours traveled by the second car before the meeting.
The first car travels for \(t+1\) hours:
\[ 50(t+1). \]The second travels
\[ 75t. \]Their distances add to \(300\):
\[ 50(t+1)+75t=300. \]Solve:
\[ 125t+50=300, \] \[ 125t=250, \] \[ t=2. \]Therefore, they meet
\[ \boxed{2\text{ hours after the second car leaves}}. \]Finding the First Traveler's Total Time
In the previous problem, the first car travels for
\[ t+1=3 \]hours.
Its distance is
\[ 50(3)=150. \]The second car also travels
\[ 75(2)=150. \]Therefore, they meet halfway between the towns.
Moving Apart after Starting at Different Points
Suppose two objects begin \(D_0\) units apart and move farther apart in opposite directions.
After time \(t\), the separation is
\[ D=D_0+r_1t+r_2t. \]Therefore,
\[ \boxed{ D=D_0+(r_1+r_2)t }. \]Example: Increasing Separation
Two vehicles begin \(40\) miles apart and move directly away from one another at \(35\) mph and \(45\) mph.
Their separation rate is
\[ 35+45=80\text{ mph}. \]After \(2\) hours,
\[ D=40+80(2)=200. \]Therefore,
\[ \boxed{200\text{ miles apart}}. \]Same-Direction Motion with an Initial Separation
Suppose two objects move in the same direction with rates \(r_f>r_s\), and the slower object begins \(L\) units ahead.
The remaining gap after time \(t\) is
\[ \boxed{ G=L-(r_f-r_s)t }. \]At the catch point,
\[ G=0. \]Example: Remaining Gap
A runner is \(2.4\) miles ahead of a cyclist. The runner moves at \(6\) mph and the cyclist at \(18\) mph.
The cyclist gains at
\[ 18-6=12\text{ mph}. \]After \(0.1\) hour, the remaining gap is
\[ 2.4-12(0.1)=1.2. \]Therefore,
\[ \boxed{1.2\text{ miles}}. \]Position Equations
Relative-motion problems can also be solved using position functions.
If object A begins at position \(p_A\) with signed velocity \(v_A\),
\[ x_A(t)=p_A+v_At. \]If object B begins at \(p_B\),
\[ x_B(t)=p_B+v_Bt. \]They meet when
\[ \boxed{x_A(t)=x_B(t)}. \]Example: Position-Equation Method
Let one vehicle begin at position \(0\) traveling east at \(60\) mph. Another begins \(90\) miles east of it, traveling east at \(45\) mph.
Their positions are
\[ x_A=60t \]and
\[ x_B=90+45t. \]Set them equal:
\[ 60t=90+45t. \] \[ 15t=90. \] \[ t=6. \]Therefore, the faster vehicle catches the slower one after
\[ \boxed{6\text{ hours}}. \]Signed Velocity Clarifies Direction
Choose east as the positive direction.
A vehicle moving east at \(50\) mph has velocity
\[ +50. \]A vehicle moving west at \(40\) mph has velocity
\[ -40. \]The relative velocity of the first with respect to the second is
\[ 50-(-40)=90. \]The sign convention naturally produces the sum for opposing directions.
Relative Velocity as a Difference
In one dimension, the velocity of A relative to B is
\[ \boxed{v_{A/B}=v_A-v_B}. \]If the velocities have opposite signs, subtraction becomes addition of their speed magnitudes.
If they have the same sign, subtraction produces the difference in speeds.
This single formula unifies same-direction and opposing-motion cases.
Closing Speed
Closing speed is the positive rate at which a gap is shrinking.
If two objects approach one another at \(70\) mph and \(50\) mph, their closing speed is
\[ 70+50=120\text{ mph}. \]If a faster car approaches a slower car from behind at \(70\) mph and \(50\) mph, the closing speed is
\[ 70-50=20\text{ mph}. \]Separation Rate
Separation rate is the rate at which the distance between objects increases.
If two objects leave the same point in opposite directions at \(25\) mph and \(35\) mph, then the separation rate is
\[ 25+35=60\text{ mph}. \]If two objects move in the same direction at \(65\) mph and \(50\) mph, the faster one increases its lead at
\[ 65-50=15\text{ mph}. \]Meeting Problems Can Be Solved Two Ways
For two objects moving toward one another, we may either:
- add the individual distances and set their sum equal to the initial separation; or
- use the combined closing speed directly.
For example,
\[ r_1t+r_2t=D \]becomes
\[ (r_1+r_2)t=D. \]The second form is simply the first form after factoring.
Pursuit Problems Can Also Be Solved Two Ways
We may equate positions:
\[ r_ft=L+r_st, \]or use relative speed:
\[ (r_f-r_s)t=L. \]The second equation is obtained by subtracting \(r_st\) from both sides of the first.
Relative-speed shortcuts are therefore consequences of ordinary distance-rate-time algebra.
Example: Find an Unknown Faster Rate
A slower vehicle is \(48\) miles ahead and travels at \(52\) mph. A faster vehicle catches it in \(3\) hours.
Let \(r\) be the faster rate.
The required gaining speed is
\[ r-52. \]Since the \(48\)-mile lead disappears in \(3\) hours,
\[ 3(r-52)=48. \]Divide by \(3\):
\[ r-52=16. \] \[ r=68. \]Therefore,
\[ \boxed{68\text{ mph}}. \]Example: Find an Unknown Slower Rate
A car traveling at \(72\) mph catches a vehicle that began \(35\) miles ahead after \(2.5\) hours.
Let \(r\) be the slower rate.
\[ (72-r)(2.5)=35. \]Then
\[ 72-r=14. \] \[ r=58. \]Therefore,
\[ \boxed{58\text{ mph}}. \]Example: Find the Starting Separation
Two trains travel toward one another at \(48\) mph and \(62\) mph. They meet after \(2.2\) hours.
Their combined rate is
\[ 48+62=110. \]Therefore, the original separation is
\[ D=110(2.2)=242. \]Thus,
\[ \boxed{242\text{ miles}}. \]Example: Find the Separation after a Given Time
Two boats begin \(180\) miles apart and travel directly toward one another at \(30\) mph and \(45\) mph.
Their closing speed is
\[ 75\text{ mph}. \]After \(2\) hours, they have closed
\[ 75(2)=150 \]miles of the original gap.
Therefore, the remaining separation is
\[ 180-150=30. \]Thus,
\[ \boxed{30\text{ miles}}. \]Do They Meet within the Given Time?
The previous boats begin \(180\) miles apart and close at \(75\) mph.
Their meeting time is
\[ \frac{180}{75}=2.4\text{ hr}. \]Therefore, after \(2\) hours they have not yet met.
This type of comparison is useful when a problem asks whether a meeting occurs before a deadline.
Relative Motion with Walking, Running, and Cycling
The same equations apply regardless of the moving objects.
A walker at \(4\) mph and a cyclist at \(16\) mph moving in the same direction have relative speed
\[ 16-4=12\text{ mph}. \]If they move in opposite directions, their separation changes at
\[ 16+4=20\text{ mph}. \]The algebra depends on motion geometry, not on the type of traveler.
Relative Motion on a Track
Circular-track problems can also use relative rates.
If two runners move in the same direction around a track of length \(L\), the faster runner gains one full lap when the relative distance gained is
\[ L. \]Therefore,
\[ \boxed{ t=\frac{L}{r_f-r_s} }. \]If they move in opposite directions, the meeting rate is \(r_1+r_2\).
Example: Same-Direction Track Problem
A circular track is \(0.5\) mile long. Two runners begin together and move in the same direction at \(8\) mph and \(6\) mph.
The faster runner gains at
\[ 8-6=2\text{ mph}. \]To gain one complete lap,
\[ t=\frac{0.5}{2}=0.25\text{ hr}. \]Therefore,
\[ \boxed{15\text{ minutes}}. \]Example: Opposite-Direction Track Problem
Two runners begin together on a \(0.6\)-mile circular track and move in opposite directions at \(7\) mph and \(5\) mph.
Their combined rate is
\[ 7+5=12\text{ mph}. \]They first meet again after their combined traveled distance equals one lap:
\[ t=\frac{0.6}{12}=0.05\text{ hr}. \]Convert to minutes:
\[ 0.05(60)=3. \]Therefore,
\[ \boxed{3\text{ minutes}}. \]Relative Motion and Unit Conversion
Relative rates must use compatible units.
If one object moves at \(20\) meters per second and another at \(54\) kilometers per hour, convert before adding or subtracting.
Since
\[ 54\text{ km/h}=15\text{ m/s}, \]the same-direction relative speed is
\[ 20-15=5\text{ m/s}. \]Unit conversion must come before relative-rate arithmetic.
Checking with the Individual Distances
Relative-speed methods are efficient, but the strongest verification is to reconstruct the original motion.
If two cars \(330\) miles apart meet in \(2.75\) hours at \(55\) mph and \(65\) mph, verify:
\[ 55(2.75)+65(2.75)=330. \]Since
\[ 151.25+178.75=330, \]the meeting-time result is confirmed.
Checking a Pursuit Solution
Suppose a car catches a truck after \(2\) hours when the truck began \(30\) miles ahead, with rates \(65\) mph and \(50\) mph.
The car travels
\[ 65(2)=130 \]miles.
The truck's position relative to the car's starting point is
\[ 30+50(2)=130. \]Their positions match, so the pursuit calculation is correct.
Common Error: Adding Speeds in Every Problem
Speeds add when the objects' motions make the gap change in opposite directional contributions, such as moving toward one another or directly apart.
In a same-direction pursuit problem, the faster object gains only at the difference:
\[ \boxed{r_f-r_s}. \]Always ask how the separation is changing.
Common Error: Subtracting Speeds in Every Problem
If two objects move toward one another, subtracting speeds understates how quickly the gap closes.
For rates \(40\) mph and \(55\) mph, the closing speed is
\[ 40+55=95\text{ mph}, \]not \(15\) mph.
Common Error: Giving Both Objects the Same Travel Time
If one object leaves earlier, the travel times are different.
If the head start is \(1.5\) hours and the later object travels for \(t\) hours, the earlier object travels for
\[ \boxed{t+1.5}. \]Using \(t\) for both loses the head-start information.
Common Error: Forgetting the Initial Lead
A pursuit problem is not solved by setting
\[ r_ft=r_st. \]If the slower object begins \(L\) units ahead, its position is
\[ L+r_st. \]The correct catch equation is
\[ \boxed{r_ft=L+r_st}. \]Common Error: Adding Distances in a Pursuit Problem
When two objects move toward one another, their traveled distances add to the original separation.
In a same-direction pursuit problem, both objects move toward the same location.
At the catch point, their positions are equal rather than their traveled distances adding to the initial lead.
Common Error: Mixing Speed and Velocity
Speed is nonnegative, while velocity includes direction.
If direction is represented with signs, a westward velocity might be
\[ -40\text{ mph}. \]The relative-velocity formula
\[ v_A-v_B \]must use signed velocities consistently.
Common Error: Ignoring Whether the Catch Is Possible
If the pursuer is not faster than the object ahead, a catch cannot occur under constant same-direction motion.
If
\[ r_f\le r_s, \]then
\[ r_f-r_s\le0. \]The gap does not shrink to zero.
Common Error: Ignoring Units
Rates cannot be meaningfully added or subtracted until their units match.
For example,
\[ 60\text{ mph} \]and
\[ 20\text{ m/s} \]must first be converted to a common unit system.
A Reliable Procedure for Opposing-Motion Problems
- Identify the initial separation.
- Record each object's rate and travel time.
- Convert units if needed.
- Determine whether the objects move toward one another or apart.
- Add the relevant speed magnitudes for the relative rate.
- Use \(D=(r_1+r_2)t\) when the times are equal.
- If start times differ, write separate distance expressions.
- Solve and reconstruct both traveled distances to verify.
A Reliable Procedure for Pursuit Problems
- Identify the slower and faster objects.
- Determine the initial lead in distance or time.
- If needed, convert a time head start into a distance lead.
- Find the gaining rate \(r_f-r_s\).
- Use \(t=L/(r_f-r_s)\) when a distance lead is known.
- Alternatively, write both position equations and set them equal.
- Verify that both objects occupy the same position at the solved time.
- Confirm that the faster rate is actually greater than the slower rate.
A Reliable Procedure Using Position Functions
- Choose an origin and positive direction.
- Assign signed velocities.
- Write \(x_A=p_A+v_At\).
- Write \(x_B=p_B+v_Bt\).
- Set the positions equal for a meeting or catch.
- Solve for \(t\).
- Substitute back to find the meeting position.
- Interpret the signs and units in context.
Practice 1: Moving Apart
Two cyclists leave the same point in opposite directions at \(16\) mph and \(21\) mph. How far apart are they after \(2.5\) hours?
Practice 2: Meeting Time
Two towns are \(420\) miles apart. Two cars leave at the same time and travel toward one another at \(60\) mph and \(80\) mph. When do they meet?
Practice 3: Meeting Location
Using Practice 2, how far from the first town does the \(60\)-mph car travel before the meeting?
Practice 4: Same-Direction Separation
Two vehicles begin side by side and travel in the same direction at \(72\) mph and \(58\) mph. How far apart are they after \(3\) hours?
Practice 5: Pursuit with a Distance Lead
A truck is \(42\) miles ahead of a car. The truck travels at \(48\) mph and the car at \(62\) mph. How long does the car take to catch the truck?
Practice 6: Pursuit with a Time Head Start
A bus travels at \(50\) mph. Ninety minutes later, a car leaves the same point in the same direction at \(70\) mph. How long after the car leaves does it catch the bus?
Practice 7: Unknown Opposing Rate
Two vehicles are \(360\) miles apart and move toward one another. One travels at \(55\) mph. They meet after \(3\) hours. Find the other rate.
Practice 8: Unknown Pursuit Rate
A slower vehicle travels at \(44\) mph and begins \(36\) miles ahead. A faster vehicle catches it in \(2\) hours. Find the faster rate.
Practice 9: Different Start Times
Two towns are \(275\) miles apart. A vehicle leaves the first town at \(45\) mph. One hour later, a second vehicle leaves the other town toward the first at \(70\) mph. How long after the second vehicle leaves do they meet?
Practice 10: Increasing Separation
Two vehicles begin \(25\) miles apart and move directly away from one another at \(30\) mph and \(50\) mph. How far apart are they after \(1.75\) hours?
Practice 11: Remaining Gap
Two vehicles begin \(150\) miles apart and move toward one another at \(40\) mph and \(35\) mph. How far apart are they after \(1.5\) hours?
Practice 12: Same-Direction Track
Two runners begin together on a \(0.4\)-mile track and run in the same direction at \(9\) mph and \(7\) mph. How long until the faster runner gains one full lap?
Practice 13: Opposite-Direction Track
Two runners begin together on a \(0.5\)-mile track and run in opposite directions at \(6\) mph and \(4\) mph. How long until they first meet again?
Practice 14: Unit Conversion
One object moves at \(25\text{ m/s}\). Another moves in the same direction at \(54\text{ km/h}\). Find their relative speed in meters per second.
Practice 15: Identify the Error
Two cars move toward one another at \(55\) mph and \(65\) mph. A student says their closing speed is
\[ 65-55=10\text{ mph}. \]Explain the error and give the correct closing speed.
Practice Answers
Practice 1
Separation rate:
\[ 16+21=37\text{ mph}. \]After \(2.5\) hours:
\[ 37(2.5)=92.5. \]Therefore,
\[ \boxed{92.5\text{ miles}}. \]Practice 2
Closing speed:
\[ 60+80=140\text{ mph}. \]Therefore,
\[ t=\frac{420}{140}=3. \] \[ \boxed{3\text{ hours}}. \]Practice 3
\[ d=60(3)=180. \]Therefore, the meeting point is
\[ \boxed{180\text{ miles from the first town}}. \]Practice 4
Relative speed:
\[ 72-58=14\text{ mph}. \]Separation after \(3\) hours:
\[ 14(3)=42. \]Therefore,
\[ \boxed{42\text{ miles}}. \]Practice 5
Gaining rate:
\[ 62-48=14\text{ mph}. \]Catch-up time:
\[ t=\frac{42}{14}=3. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 6
Convert the head start:
\[ 90\text{ min}=1.5\text{ hr}. \]The bus travels
\[ 50(1.5)=75 \]miles before the car starts.
The gaining rate is
\[ 70-50=20. \]Therefore,
\[ t=\frac{75}{20}=3.75. \] \[ \boxed{3.75\text{ hours}}. \]Practice 7
Combined speed must satisfy
\[ (55+r)(3)=360. \] \[ 55+r=120. \] \[ r=65. \]Therefore,
\[ \boxed{65\text{ mph}}. \]Practice 8
The required gaining speed is
\[ \frac{36}{2}=18\text{ mph}. \]Therefore,
\[ r-44=18. \] \[ r=62. \]Thus,
\[ \boxed{62\text{ mph}}. \]Practice 9
Let \(t\) be the second vehicle's travel time.
\[ 45(t+1)+70t=275. \] \[ 115t+45=275. \] \[ 115t=230. \] \[ t=2. \]Therefore,
\[ \boxed{2\text{ hours}}. \]Practice 10
Separation rate:
\[ 30+50=80. \]Increase in separation:
\[ 80(1.75)=140. \]Add the original \(25\)-mile separation:
\[ 25+140=165. \]Therefore,
\[ \boxed{165\text{ miles}}. \]Practice 11
Closing speed:
\[ 40+35=75. \]Distance closed:
\[ 75(1.5)=112.5. \]Remaining gap:
\[ 150-112.5=37.5. \]Therefore,
\[ \boxed{37.5\text{ miles}}. \]Practice 12
Relative speed:
\[ 9-7=2\text{ mph}. \]Time to gain \(0.4\) mile:
\[ t=\frac{0.4}{2}=0.2\text{ hr}. \]Convert to minutes:
\[ 0.2(60)=12. \]Therefore,
\[ \boxed{12\text{ minutes}}. \]Practice 13
Combined rate:
\[ 6+4=10\text{ mph}. \]Time for one combined lap:
\[ t=\frac{0.5}{10}=0.05\text{ hr}. \]Convert to minutes:
\[ 0.05(60)=3. \]Therefore,
\[ \boxed{3\text{ minutes}}. \]Practice 14
Convert
\[ 54\text{ km/h}=15\text{ m/s}. \]The same-direction relative speed is
\[ 25-15=10. \]Therefore,
\[ \boxed{10\text{ m/s}}. \]Practice 15
Because the cars move toward one another, each car reduces the gap. Their speed magnitudes must be added:
\[ 55+65=120. \]Therefore, the correct closing speed is
\[ \boxed{120\text{ mph}}. \]What to Remember
Relative motion focuses on how quickly the distance between moving objects changes.
For opposite-direction motion,
\[ \boxed{r_{\text{relative}}=r_1+r_2}. \]For same-direction motion,
\[ \boxed{r_{\text{relative}}=|r_1-r_2|}. \]For two objects moving toward one another from initial separation \(D\),
\[ \boxed{ t=\frac{D}{r_1+r_2} }. \]For a faster object pursuing a slower one with initial lead \(L\),
\[ \boxed{ t=\frac{L}{r_f-r_s} }. \]The most important habit is to identify how the gap is changing before choosing whether speeds should be added or subtracted.
Where We Go Next
Motion problems combine rates through distance. Another major applied problem family combines rates through completed work. Jobs, machines, pipes, crews, and other productive processes can all be modeled by asking what fraction of a task is completed per unit of time.
The next lesson, Work, Rate, and Time, develops that framework.