Question

The distance between towns A and B is 360km.a minibus left town A at 8.15a.m and traveled towards town B at an average speed of 90km/hr. A matatu left town B two and a third hours later on the same day and traveled towards A at average speed of 110km/hr.
(a) (I)at what time did the two vehicles meet?
(Ii)how far from A did the two vehicles meet?
(b) a motorist started from his home at 10.30 a.m.on the same day as the matatu And traveled at an average speed of 100km/hr. He arrives at B at the same time as the minibus. Calculate that distance from A to his house

Answer 1

To solve this problem, we'll break it down into parts (a) and (b) as specified.

### Part (a)

**(I)** Let's denote:

- Distance between Town A and B: \( D = 360 \) km
- Speed of the minibus: \( v_A = 90 \) km/hr
- Speed of the matatu: \( v_B = 110 \) km/hr

1. **Time taken by the minibus to travel until meeting:**

The minibus leaves Town A at 8:15 AM. Let's denote the time taken by it until the meeting point as \( t \) hours.

Distance covered by the minibus at the time they meet:
\[
\text{Distance} = v_A \cdot t = 90t \text{ km}
\]

2. **Travel time of the matatu:**

The matatu leaves Town B two and a third hours later. Two and a third hours is equal to \( \frac{7}{3} \) hours, which is approximately 2 hours and 20 minutes. Thus, the matatu leaves Town B at:
\[
8:15 \, \text{AM} + 2\frac{1}{3} \, \text{hours} = 10:35 \, \text{AM}
\]
The matatu travels for \( t - \frac{7}{3} \) hours until they meet.

Therefore, the distance covered by the matatu at the time they meet:
\[
\text{Distance} = v_B \cdot \left(t - \frac{7}{3}\right) = 110\left(t - \frac{7}{3}\right) \text{ km}
\]

3. **Setting up the equation:**

The total distance covered by both vehicles is equal to the distance between A and B:
\[
90t + 110\left(t - \frac{7}{3}\right) = 360
\]
Expanding the equation:
\[
90t + 110t - \frac{770}{3} = 360
\]
Combining like terms:
\[
200t - \frac{770}{3} = 360
\]
To eliminate the fraction, multiply the entire equation by 3:
\[
600t - 770 = 1080
\]
Adding 770 to both sides:
\[
600t = 1850
\]
Dividing by 600:
\[
t = \frac{1850}{600} = \frac{37}{12} \text{ hours} = 3\frac{1}{12} \text{ hours}
\]

Now convert this into a time format:
\[
3\frac{1}{12} \text{ hours} = 3 \text{ hours and } 5 \text{ minutes}
\]
Therefore, the minibus meets the matatu at:
\[
8:15 \, \text{AM} + 3 \text{ hours } 5 \text{ minutes} = 11:20 \, \text{AM}
\]

**(II)** To find how far from A they meet, we can plug in the time \( t \) into the distance equation of the minibus:
\[
\text{Distance from A} = 90t = 90 \times \frac{37}{12} = 277.5 \text{ km}
\]

### Part (b)

The motorist leaves at 10:30 AM and arrives at Town B at the same time as the minibus.

1. **Time taken by the minibus from Town A to Town B:**
The minibus left at 8:15 AM and meets at 11:20 AM:
\[
\text{Time taken} = 11:20 \, \text{AM} - 8:15 \, \text{AM} = 3 \text{ hours } 5 \text{ minutes} \text{ (or } \frac{37}{12} \text{ hours )}
\]

2. Since the motorist arrives at B at the same time as the minibus after leaving at 10:30 AM, we calculate the time he took:
\[
\text{Time for the motorist} = 11:20 \, \text{AM} - 10:30 \, \text{AM} = 50 \text{ minutes} = \frac{5}{6} \text{ hours}
\]

3. **Distance traveled by the motorist:**
The motorist travels at 100 km/hr:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 100 \cdot \frac{5}{6} = \frac{500}{6} \approx 83.33 \text{ km}
\]

Thus, the distance from A to the motorist's house is approximately **83.33 km**.