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.

Question

(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.