Meeting Problems (Opposite Directions)
When two objects move towards each other, the distance between them shrinks very quickly. To simplify the math, we calculate their Relative Speed. We pretend one object is completely stationary, and the other object is doing all the moving.
The "Meeting" Rule
When objects move in opposite directions (towards each other), you ADD their speeds together to find the Relative Speed.
Why do we ADD the speeds?
In one hour, Car A covers 60 km towards the center, and Car B covers 40 km towards the center. Because they are working together to eat up the distance, the total ground covered in that hour is $60 + 40 = 100 \text{ km}$.
Example: Two Trains
"Train A and Train B are $300 \text{ km}$ apart. Train A drives East at $60 \text{ km/h}$. Train B drives West at $40 \text{ km/h}$. How long until they meet?"
1. Find Relative Speed:
$$ 60 + 40 = 100 \text{ km/h} $$(Pretend Train B is parked, and Train A is moving at $100 \text{ km/h}$ towards it).
2. Calculate Time:
$$ t = \frac{300 \text{ km}}{100 \text{ km/h}} = 3 \text{ hours} $$Knowledge Check 1
Marking Schedule:
Step 1: Relative Speed $= 80 + 70 = 150 \text{ km/h}$.
Step 2: Time $= 450 \div 150 = 3 \text{ hours}$.
Chasing Problems (Same Direction)
When two objects move in the same direction, the faster object slowly gains ground on the slower one. Again, we pretend the slower object is completely stopped, and the faster object is closing the gap using its "net" speed advantage.
The "Chasing" Rule
When objects move in the same direction, you SUBTRACT their speeds to find the Relative Speed (the rate at which the gap actually closes).
Why do we SUBTRACT the speeds?
The distance traveled by the faster object is partially cancelled out by the slower object running away. The Police Officer covers 15 km, but because the Thief ran 12 km forward at the same time, the officer only gains $15 - 12 = 3 \text{ km}$ of actual ground.
Example: Police Chase
"A thief runs at $12 \text{ km/h}$. A police officer chases him at $15 \text{ km/h}$. The thief has a $6 \text{ km}$ head start. How long until the arrest?"
1. Find Relative Speed:
$$ 15 - 12 = 3 \text{ km/h} $$(Pretend the thief is standing still, and the police officer is closing the gap at $3 \text{ km/h}$).
2. Calculate Time:
$$ t = \frac{6 \text{ km}}{3 \text{ km/h}} = 2 \text{ hours} $$Knowledge Check 2
Marking Schedule:
Q1: Relative Speed $= 8 - 5 = 3 \text{ m/s}$. Time $= 24 \div 3 = 8 \text{ seconds}$.
Q2: Relative Speed $= 100 - 80 = 20 \text{ km/h}$. Time $= 20 \div 20 = 1 \text{ hour}$.
Challenge: Time-Shifted Departures
In advanced problems, the "head start" is rarely given as a clean distance. Instead, you are told that one object left earlier than the other. You must manually calculate the head start distance before you can use the relative speed formula.
Pro-Tip: Draw a Diagram!
For time-shifted problems, always sketch a quick timeline or distance line. Mark exactly where Object A is at the exact moment Object B finally starts moving. Visualizing that physical "Head Start" gap makes the algebra much less confusing.
Target Problem Analysis
"Car A leaves town at 1:00 PM travelling at 50 km/h. Car B leaves the same town at 3:00 PM travelling at 70 km/h, chasing Car A. How many hours will it take for Car B to catch Car A?"
Car A left 2 hours before Car B started moving.
In those 2 hours, Car A was driving at $50 \text{ km/h}$.
$$ \text{Head Start} = 2 \text{ hours} \times 50 \text{ km/h} = 100 \text{ km} $$
Relative Speed $= 70 - 50 = 20 \text{ km/h}$.
$$ \text{Time} = \frac{100 \text{ km}}{20 \text{ km/h}} = 5 \text{ hours} $$
Knowledge Check 3
A cyclist leaves at 8:00 AM travelling at $20 \text{ km/h}$. A car leaves from the same spot at 8:30 AM travelling at $60 \text{ km/h}$ to catch him.
Marking Schedule:
Q1: Time gap is 30 mins ($0.5 \text{ hours}$). Head start $= 0.5 \times 20 = 10 \text{ km}$.
Q2: Relative speed $= 60 - 20 = 40 \text{ km/h}$. Time to catch $= 10 \div 40 = 0.25 \text{ hours}$ (or 15 mins).
Synthesis Task
Combine all skills. Complete 18 mixed-context questions. You have 2 attempts. (Use exact decimals).
Section A: Base Meeting Problems (Sum)
Section B: Base Chasing Problems (Difference)
Section C: Time-Shifted Departures
Section D: Advanced Challenge Variations
Student Mastery Report
Take a screenshot of this report to submit to Mr. Shen.