A truck left Pretoria at 09:00 and travelled to Benoni at a speed of 50km/h throughout

the whole journey. A car left Benoni at 09:30 and travelled to Pretoria. The speed of

the car remained the same until it passed the lorry at 12:00. The truck has travelled

3

7 of the journey when it passed the car. The speed of the car decreased by 5km/h

after it passed the truck and it travelled at this new speed for the rest of the journey.

At what time did the car reach Pretoria 

To solve this problem, let's break it down step by step and assign some variables. We know that the truck and the car passed each other at 12:00, and the truck had traveled \( \frac{3}{7} \) of the journey at that point. Here's how we can set it up:

### 1. Let the total distance between Pretoria and Benoni be \( d \) km.

### 2. Truck's Travel:
- The truck started at 09:00 and traveled at a constant speed of 50 km/h.
- By the time the truck and car passed each other at 12:00, the truck had been traveling for 3 hours.
- Since the truck had traveled \( \frac{3}{7} \) of the total distance, we can calculate the distance traveled by the truck as:
\[
\text{Distance traveled by the truck} = \frac{3}{7}d
\]
- The truck's distance formula is:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
So in 3 hours:
\[
\frac{3}{7}d = 50 \times 3
\]
\[
\frac{3}{7}d = 150
\]
Solving for \( d \):
\[
d = \frac{150 \times 7}{3} = 350 \text{ km}
\]

The total distance between Pretoria and Benoni is 350 km.

### 3. Car's Travel:
- The car started at 09:30 and passed the truck at 12:00, meaning it had been traveling for 2.5 hours at a constant speed.
- During this time, the car traveled the remaining \( \frac{4}{7} \) of the total journey (because the truck had already covered \( \frac{3}{7} \) of it). The car's speed was \( v \) km/h, and in 2.5 hours, the car traveled:
\[
\frac{4}{7}d = v \times 2.5
\]
Substitute \( d = 350 \) km:
\[
\frac{4}{7} \times 350 = v \times 2.5
\]
\[
200 = v \times 2.5
\]
Solving for \( v \):
\[
v = \frac{200}{2.5} = 80 \text{ km/h}
\]
So the car's initial speed was 80 km/h until it passed the truck at 12:00.

### 4. Speed Decrease After Passing the Truck:
- After passing the truck, the car's speed decreased by 5 km/h, so its new speed was:
\[
80 - 5 = 75 \text{ km/h}
\]
- The car still had \( \frac{3}{7} \) of the journey left after passing the truck. The distance remaining is:
\[
\frac{3}{7} \times 350 = 150 \text{ km}
\]
- The time taken to cover the remaining distance at 75 km/h is:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{150}{75} = 2 \text{ hours}
\]

### 5. Arrival Time:
- Since the car passed the truck at 12:00 and took 2 more hours to cover the remaining distance, the car arrived in Pretoria at:
\[
12:00 + 2 \text{ hours} = 14:00
\]

Thus, the car reached Pretoria at **14:00**.

---

Would you like further details or have any questions? Here are some related questions:

1. How would the result change if the truck traveled at 60 km/h instead?
2. What if the car reduced its speed by 10 km/h instead of 5 km/h after passing the truck?
3. How can we model the meeting point between the truck and the car mathematically?
4. What if the journey between Pretoria and Benoni was 400 km instead of 350 km?
5. How would we calculate the car's total travel time?

**Tip:** When solving such problems, always break them into smaller parts and deal with one section at a time, such as determining distances, speeds, and times.

A truck left Pretoria at 09:00 and travelled to Benoni at a speed of 50km/h throughout the whole journey. A car left Benoni at 09:30 and travelled to Pretoria. The speed of the car remained the same until it passed the lorry at 12:00. The truck has travelled 3/7 of the journey when it passed the car. The speed of the car decreased by 5km/h after it passed the truck and it travelled at this new speed for the rest of the journey. At what time did the car reach Pretoria?

Solve a real-world travel time problem involving a truck and a car traveling between Pretoria and Benoni. Learn how to calculate the time the car reaches its destination using algebraic concepts of distance, speed, and time, including proportional relationships. Suitable for Grades 10-12.

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