To solve this problem, we need to determine when the car will catch up with the bus. Let's break it down step by step, using a figure and equations.

Step 1: Define variables and relationships

Let:

• $t$ be the time (in hours) that the car takes to catch up with the bus.
• The bus travels at a speed of $80 \, \text{km/h}$.
• The car travels at a speed of $120 \, \text{km/h}$.

The key point is that the bus has a 2-hour head start, so in that time, the bus will have covered some distance before the car starts.

Step 2: Calculate the head start distance of the bus

In the first 2 hours, the bus will travel a distance: $\text{Distance of the bus in 2 hours} = 80 \, \text{km/h} \times 2 \, \text{hours} = 160 \, \text{km}$ So, the bus will be 160 km ahead of the car when the car starts.

Step 3: Set up the relative distance equation

Now, after the car starts, the distance covered by the bus and the car can be described as:

• Distance covered by the bus after the car starts: $80 \times t$
• Distance covered by the car: $120 \times t$

Since the car will catch up with the bus, we need the total distance covered by the car to equal the distance covered by the bus plus the initial head start of 160 km. Therefore, we set up the equation: $120t = 80t + 160$

Step 4: Solve the equation

Now, solve for $t$:

$120t - 80t = 160$ $40t = 160$ $t = \frac{160}{40} = 4 \, \text{hours}$

Conclusion:

The car will catch up with the bus after 4 hours.

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If you would like me to include a diagram, please let me know! Would you like any further details or explanations?

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Here are 5 related questions to expand this understanding:

1. What is the total distance traveled by the car and bus when the car catches up?
2. If the bus increased its speed to 90 km/h, how long would it take the car to catch up?
3. What if the car had a delay of 3 hours instead of 2, how long would it take to catch up?
4. Can you solve this using the concept of relative speed directly?
5. What if both vehicles were moving in opposite directions, when would they meet?

Tip: Whenever two moving objects are involved, using the relative speed approach can often simplify the calculations.