Special Systems and Applications

Interpreting the graph

• The slope of each line represents velocity.

• A positive slope represents motion in the chosen positive direction.

• A negative slope represents motion in the opposite direction.

• The vertical intercept represents initial position.

• The intersection represents the meeting time and position.

The intersection is a solution to both equations.

Worked Example: Finding Two Unknown Speeds

Two cyclists travel towards each other from towns 150 km apart.

They meet after 2 hours.

Cyclist A travels 15 km/h faster than Cyclist B.

Find their speeds.

Step 1: Define variables

x = speed of Cyclist A in km/h

y = speed of Cyclist B in km/h

Step 2: Write the speed relationship

Cyclist A travels 15 km/h faster:

x − y = 15

Step 3: Write the distance relationship

They travel for 2 hours and cover a combined distance of 150 km.

2x + 2y = 150

Divide by 2:

x + y = 75

Step 4: Solve using elimination

x − y = 15

x + y = 75

Add:

2x = 90

x = 45

Then:

y = 75 − 45

y = 30

Step 5: Interpret

Cyclist A travels at 45 km/h.

Cyclist B travels at 30 km/h.

Step 6: Verify

Difference in speed:

45 − 30 = 15 km/h

Distances:

Cyclist A: 45 × 2 = 90 km

Cyclist B: 30 × 2 = 60 km

Total:

90 + 60 = 150 km

Answer: Cyclist A = 45 km/h; Cyclist B = 30 km/h.

This example is particularly important because the unknown quantities are speeds rather than times or distances.

Worked Example: Journey at Two Different Speeds

A delivery driver travels a total of 180 km.

Part of the journey is completed at 60 km/h.

The remaining distance is completed at 90 km/h.

The total driving time is 2.5 hours.

Find the distance travelled at each speed.

Step 1: Define variables

x = distance travelled at 60 km/h, in kilometres

y = distance travelled at 90 km/h, in kilometres

Step 2: Write the distance equation

x + y = 180

Step 3: Write the time equation

Time = Distance / Speed

Time at 60 km/h:

x/60

Time at 90 km/h:

y/90

Total time:

x/60 + y/90 = 2.5

Step 4: Eliminate fractions

Multiply the second equation by 180:

3x + 2y = 450

The system is:

x + y = 180

3x + 2y = 450

Step 5: Solve

Multiply the first equation by 2:

2x + 2y = 360

Subtract:

x = 90

Then:

y = 180 − 90

y = 90

Step 6: Verify

Total distance:

90 + 90 = 180 km

Total time:

90/60 + 90/90

= 1.5 + 1

= 2.5 hours

Answer: 90 km at 60 km/h and 90 km at 90 km/h.

Important distinction

The driver spends different amounts of time travelling at the two speeds.

Time at 60 km/h = 1.5 hours

Time at 90 km/h = 1 hour

Although the distances are equal, the times are not.

Worked Example: A Boat Travelling Upstream and Downstream

A boat travels along a river.

Its speed in still water is unknown.

The river current also has an unknown speed.

The boat travels downstream at 18 km/h and upstream at 10 km/h.

Find the boat's speed in still water and the speed of the current.

Step 1: Define variables

b = boat speed in still water, in km/h

c = river current speed, in km/h

Step 2: Write equations

When travelling downstream, the current helps the boat:

b + c = 18

When travelling upstream, the current opposes the boat:

b − c = 10

Step 3: Solve using elimination

Add the equations:

(b + c) + (b − c) = 18 + 10

2b = 28

b = 14

Substitute:

14 + c = 18

c = 4

Step 4: Interpret

Boat speed in still water:

14 km/h

River current speed:

4 km/h

Step 5: Verify

Downstream speed:

14 + 4 = 18 km/h

Upstream speed:

14 − 4 = 10 km/h

Answer: Boat speed = 14 km/h; current speed = 4 km/h.

Worked Example: Round Trip with Different Speeds

A boat travels 24 km downstream and then returns 24 km upstream.

The downstream journey takes 1 hour.

The upstream journey takes 2 hours.

Find the boat's speed in still water and the river current speed.

Step 1: Calculate the travel speeds

Downstream:

v = 24/1 = 24 km/h

Upstream:

v = 24/2 = 12 km/h

Step 2: Define variables

b = boat speed in still water

c = current speed

Step 3: Construct the system

b + c = 24

b − c = 12

Step 4: Solve

Add:

2b = 36

b = 18

Then:

c = 24 − 18

c = 6

Step 5: Verify

Downstream:

18 + 6 = 24 km/h

Upstream:

18 − 6 = 12 km/h

Travel times:

24/24 = 1 hour

24/12 = 2 hours

Answer: Boat speed = 18 km/h; current speed = 6 km/h.

Worked Example: Aircraft and Wind Speed

An aircraft travels at 650 km/h relative to the air.

With a tailwind, its ground speed is 720 km/h.

What is the wind speed?

Let:

a = aircraft speed relative to the air

w = wind speed

For a tailwind:

a + w = 720

Given:

a = 650

Substitute:

650 + w = 720

w = 70

Answer: The tailwind speed is 70 km/h.

Now suppose the aircraft's speed relative to the air is unknown, but its ground speed is:

• 720 km/h with a tailwind.

• 580 km/h with an equal-strength headwind.

Then:

a + w = 720

a − w = 580

Add:

2a = 1300

a = 650

Then:

w = 70

Answer: Aircraft airspeed = 650 km/h; wind speed = 70 km/h.

This simplified model assumes the aircraft flies directly along or against a steady wind.

Worked Example: Two Runners on a Circular Track

Two runners start together at the same point on a 400 m circular track.

They run in the same direction.

Runner A runs at 6 m/s.

Runner B runs at 4 m/s.

When will Runner A first lap Runner B?

Step 1: Identify the important relationship

Runner A must travel one complete lap farther than Runner B.

Therefore:

Distance A − Distance B = 400 m

Step 2: Define variables

t = time in seconds

d₁ = distance travelled by Runner A

d₂ = distance travelled by Runner B

Step 3: Construct equations

d₁ = 6t

d₂ = 4t

d₁ − d₂ = 400

Step 4: Solve

6t − 4t = 400

2t = 400

t = 200

Step 5: Verify

Runner A travels:

6(200) = 1200 m

Runner B travels:

4(200) = 800 m

Difference:

1200 − 800 = 400 m

Answer: Runner A first laps Runner B after 200 seconds, or 3 minutes 20 seconds.

This example shows how relative speed can be used for circular motion when both runners maintain constant speeds.

Worked Example: Determining Whether Two Objects Will Meet

Two cars travel along the same straight road.

Car A starts at position 20 km and travels at 60 km/h.

Car B starts at position 100 km and travels at 80 km/h.

Both travel in the same direction.

Will Car A ever catch Car B after the start of the observation?

Step 1: Write position equations

Car A:

x = 20 + 60t

Car B:

x = 100 + 80t

Step 2: Solve the system

20 + 60t = 100 + 80t

−80 = 20t

t = −4

Step 3: Interpret

The algebraic intersection occurs at t = −4 hours.

However, t represents time after the observation begins.

Therefore, the practical domain is:

t ≥ 0

A negative meeting time lies outside the relevant domain.

Physical explanation

Car B starts ahead of Car A and travels faster.

The separation increases rather than decreases.

Answer: Car A will not catch Car B after the observation begins.

This is an important example of a mathematically valid intersection that does not represent a future meeting.

Solving Motion Problems Using Graphs

Graphs are useful because they show how motion changes over time.

For constant-velocity motion, the position-time graph is a straight line.

The general equation is:

x = x₀ + vt

This has the same mathematical structure as:

y = mx + b

The correspondence is:

Parallel position-time graphs

Suppose:

Object A:

x = 5t + 10

Object B:

x = 5t + 30

Both objects move at 5 m/s.

Their initial positions differ by 20 m.

The graphs are parallel.

The objects never meet because they maintain a constant separation of 20 m.

Coincident position-time graphs

Suppose:

Object A:

x = 5t + 10

Object B:

2x = 10t + 20

Simplify Object B:

x = 5t + 10

The equations are identical.

Both objects have the same position at every time in the model.

Their position-time graphs are coincident.

Intersecting position-time graphs

Suppose:

Object A:

x = 4t + 10

Object B:

x = 2t + 30

Set equal:

4t + 10 = 2t + 30

2t = 20

t = 10

Position:

x = 4(10) + 10

x = 50 m

The objects meet after 10 seconds at position 50 m.

Choosing the Correct Equations

Different motion problems require different relationships.

These equations are useful starting points, but the exact system must always match the situation described.

Interpreting Motion Solutions

A solution should always include the correct units.

For example, if a system produces:

t = 4

x = 240

and t is measured in hours while x is measured in kilometres, the interpretation is:

"The objects meet after 4 hours at position 240 km."

Simply writing "(4, 240)" is incomplete because it does not explain the physical meaning.

Important questions

After solving, ask:

• Does the time represent seconds, minutes, or hours?

• Is the position measured from the correct starting point?

• Is the direction correct?

• Is the time positive?

• Does the object travel a realistic distance?

• Does the solution satisfy both motion equations?

• Does the solution occur within the stated time interval?

These questions help distinguish a correct mathematical calculation from a meaningful physical answer.

Checking Whether Answers Are Reasonable

Checking travel time

If two objects start 100 km apart and approach each other at a combined speed of 50 km/h, they should meet after:

100/50 = 2 hours

An answer of 20 hours would be unreasonable.

Checking meeting position

If two objects meet between towns 200 km apart, the distances travelled towards each other should add to 200 km.

Checking speed

If a car travels 120 km in 2 hours, its average speed is:

120/2 = 60 km/h

A result of 600 km/h would suggest an error in the calculation or units.

Checking negative values

Negative position can be valid if the coordinate system allows it.

Negative velocity can also be valid because it indicates direction.

However, negative elapsed time is generally not valid when the question asks about events after the starting moment.

Checking assumptions

Most introductory motion systems assume constant speeds or velocities.

In real life, vehicles accelerate, decelerate, stop, and change direction.

The model is useful only when its assumptions are reasonable for the situation.

Real-World Applications

Transport planning

Travel equations can estimate meeting times, journey durations, and distances.

They can help compare routes and travel speeds.

Logistics and delivery

Businesses use motion models to estimate delivery times, vehicle positions, and travel schedules.

Sports

Relative speed can be used to study runners, cyclists, and racing vehicles.

Aviation

Aircraft ground speed depends on airspeed and wind velocity.

Marine transport

Boat speed relative to the ground depends on the boat's speed through water and the river current.

Physics and engineering

Position-time equations are used to model moving objects and understand their relative motion.

More advanced models include acceleration and nonlinear motion.

Common Mistakes

Confusing distance with displacement

Distance is the total path length, while displacement includes direction.

Confusing speed with velocity

Speed is scalar; velocity includes direction.

Using inconsistent units

For example, multiplying km/h by seconds without converting time will produce an incorrect result.

Forgetting a head start

If one object begins earlier, it may already be some distance ahead when the second starts.

Adding speeds when they should be subtracted

For objects moving towards each other, closing speed is usually the sum of speeds.

For objects moving in the same direction, closing speed is the difference.

Using the wrong sign for velocity

Motion in the negative direction should have negative velocity in the position equation.

Confusing meeting time with total travel time

An object that starts earlier has travelled longer by the time the objects meet.

Assuming every algebraic intersection represents a future meeting

A negative meeting time may fall outside the problem's domain.

Ignoring initial positions

The equation x = vt applies when the initial position is zero.

Otherwise, use x = x₀ + vt.

Failing to verify

Always substitute the answer into both original motion equations.

Key Terms

• Motion: A change in position over time.

• Distance: The total length of the path travelled.

• Displacement: The change in position, including direction.

• Speed: Distance travelled per unit time.

• Velocity: Rate of change of displacement.

• Average speed: Total distance divided by total time.

• Constant speed: Motion in which speed does not change.

• Constant velocity: Motion with unchanged speed and direction.

• Position: Location relative to a chosen reference point.

• Initial position: Position at the starting time.

• Relative speed: The rate at which separation between objects changes.

• Closing speed: The rate at which two objects approach each other.

• Meeting point: The position where two objects occupy the same location at the same time.

• Catch-up time: The time required for a faster object to reach a slower object.

• Position-time graph: A graph showing position as a function of time.

• Slope: The gradient of a line; velocity on a position-time graph.

• System of equations: Two or more equations solved simultaneously.

• Substitution: Solving by replacing one variable with an equivalent expression.

• Elimination: Solving by combining equations to remove a variable.

• Domain: The set of allowed input values.

• Verification: Checking that the answer satisfies the original conditions.

Key Takeaways

• Motion problems can be represented using systems of linear equations.

• The fundamental relationship is d = vt.

• Constant-velocity position equations take the form x = x₀ + vt.

• Variables must be clearly defined with appropriate units.

• Objects moving towards each other have a closing speed equal to the sum of their speeds.

• Objects moving in the same direction may have a closing speed equal to the difference between their speeds.

• A meeting point occurs when two objects have the same position at the same time.

• Position-time graphs allow motion systems to be interpreted visually.

• The slope of a position-time graph represents velocity.

• Substitution and elimination can determine unknown speeds, times, and positions.

• Solutions must be interpreted using the original context.

• Every answer should be verified for mathematical correctness and physical reasonableness.

Check Your Understanding

Recall and Understanding

1. State the relationship between distance, speed, and time.

2. Explain the difference between distance and displacement.

3. Why must units be consistent when solving motion problems?

4. What does the intersection of two position-time graphs represent?

5. Explain the difference between relative speed for objects moving towards each other and objects moving in the same direction.

Application

6. Two cyclists start 120 km apart and travel towards each other at 25 km/h and 35 km/h. How long will it take them to meet, and how far will each cyclist travel?

7. A car travels at 70 km/h. Another car starts from the same location 2 hours later at 105 km/h. How long after the second car starts will it catch the first car?

8. Two trains start 360 km apart and travel towards each other at 80 km/h and 100 km/h. Determine their meeting time and the distance travelled by each train.

9. A boat travels downstream at 24 km/h and upstream at 16 km/h. Find the boat's speed in still water and the river current speed.

10. A driver travels 240 km. Part of the journey is completed at 60 km/h and the remainder at 80 km/h. The total driving time is 3.5 hours. Find the distance travelled at each speed.

Analysis and Challenge

11. Two runners start at the same point on a 400 m circular track and run in the same direction at 7 m/s and 5 m/s. When will the faster runner first lap the slower runner?

12. Two objects move according to:

x = 15 + 6t

x = 75 − 4t

Find the meeting time and position.

13. Two vehicles travel in the same direction:

Vehicle A: x = 20 + 50t

Vehicle B: x = 100 + 70t

Determine whether Vehicle A catches Vehicle B for t ≥ 0.

14. Two cyclists travel towards each other from towns 210 km apart. They meet after 3 hours. One cyclist travels 10 km/h faster than the other. Find their speeds.

15. An aircraft travels at 760 km/h with a tailwind and 640 km/h against a headwind of the same magnitude. Find its speed relative to the air and the wind speed.

Answers to Check Your Understanding

1.

Distance = Speed × Time

d = vt

2. Distance is the total path length travelled. Displacement is the change in position and includes direction.

3. Units must be consistent so that the calculated quantities have the correct dimensions. For example, km/h multiplied by hours gives kilometres.

4. The intersection represents the time and position at which two objects occupy the same location.

5. When objects move towards each other, the closing speed is the sum of their speeds. When they move in the same direction, the closing speed is the difference between their speeds, provided the faster object is behind.

6.

Combined speed:

25 + 35 = 60 km/h

Meeting time:

t = 120/60 = 2 hours

Distances:

Cyclist A:

25(2) = 50 km

Cyclist B:

35(2) = 70 km

Verification:

50 + 70 = 120 km

Answer: They meet after 2 hours. The cyclists travel 50 km and 70 km.

7.

The first car's head start is:

70 × 2 = 140 km

Let t = time after the second car starts.

First car position:

d = 70t + 140

Second car position:

d = 105t

Set equal:

70t + 140 = 105t

140 = 35t

t = 4

Answer: The second car catches the first after 4 hours, or 6 hours after the first car departed.

Verification:

First car: 70 × 6 = 420 km

Second car: 105 × 4 = 420 km

8.

Combined speed:

80 + 100 = 180 km/h

Meeting time:

t = 360/180 = 2 hours

Train A distance:

80 × 2 = 160 km

Train B distance:

100 × 2 = 200 km

Answer: They meet after 2 hours, having travelled 160 km and 200 km.

9.

Let b = boat speed in still water and c = current speed.

b + c = 24

b − c = 16

Add:

2b = 40

b = 20

Then:

c = 4

Answer: Boat speed = 20 km/h; current speed = 4 km/h.

10.

Let x = distance at 60 km/h and y = distance at 80 km/h.

x + y = 240

x/60 + y/80 = 3.5

Multiply the second equation by 240:

4x + 3y = 840

Multiply the first equation by 3:

3x + 3y = 720

Subtract:

x = 120

Then:

y = 120

Verification:

120/60 + 120/80

= 2 + 1.5

= 3.5 hours

Answer: 120 km at each speed.

11.

Relative speed:

7 − 5 = 2 m/s

The faster runner must gain one full lap:

400 m

Time:

t = 400/2

t = 200 seconds

Answer: 200 seconds, or 3 minutes 20 seconds.

12.

Set positions equal:

15 + 6t = 75 − 4t

10t = 60

t = 6 seconds

Position:

x = 15 + 6(6)

x = 51 m

Verification:

75 − 4(6) = 51 m

Answer: They meet after 6 seconds at position 51 m.

13.

Set positions equal:

20 + 50t = 100 + 70t

−80 = 20t

t = −4

The solution lies outside the domain t ≥ 0.

Vehicle B starts ahead and moves faster.

Answer: Vehicle A never catches Vehicle B after the observation begins.

14.

Let x = faster speed and y = slower speed.

Speed difference:

x − y = 10

Combined distance:

3x + 3y = 210

Divide by 3:

x + y = 70

Add equations:

2x = 80

x = 40

Then:

y = 30

Verification:

40 − 30 = 10 km/h

3(40) + 3(30) = 120 + 90 = 210 km

Answer: The cyclists travel at 40 km/h and 30 km/h.

15.

Let a = aircraft speed relative to the air and w = wind speed.

With tailwind:

a + w = 760

Against headwind:

a − w = 640

Add:

2a = 1400

a = 700

Substitute:

700 + w = 760

w = 60

Verification:

700 + 60 = 760 km/h

700 − 60 = 640 km/h

Answer: Aircraft airspeed = 700 km/h; wind speed = 60 km/h.

Final Summary

Distance, speed, and motion problems demonstrate how systems of equations can model moving objects and predict when and where they meet.

By combining the relationship d = vt with position equations and information about direction, starting positions, and travel times, we can solve a wide variety of practical problems.

These methods are useful in transport, logistics, sports, aviation, marine travel, and physics.

The central principle is that two objects meet when they occupy the same position at the same time. Systems of equations allow us to determine exactly when and where that happens, provided the model's assumptions are satisfied.