KCSE 2025 Maths P1 Q17 — Speed, Distance & Time (Meeting Point & Average Speed)

KCSE 2025 Form 4 Measurement

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The Question

“The road connecting town A and town B is 160 km long. A lorry travelling at an average speed of 45 km/h left town A for town B at 11:50 a.m. At the same time a car travelling at an average speed of 75 km/h left town B for town A. The two vehicles met at C, a town along the road. (a) Determine (i) the time when the two vehicles met, and (ii) the distance in km from town A to town C. (b) The car stopped at town C for 1 hour 40 minutes. The lorry continued to B at 45 km/h. After the stop the car left C for A and arrived at the same time the lorry reached B. Determine the average speed of the car from town C to town A.”

1

Find the closing speed

Because the two vehicles travel towards each other, the gap between them shrinks at the sum of their speeds. Adding the lorry's speed and the car's speed gives how fast the 160 km gap is closing.

Closing speed=45+75=120 km/h\text{Closing speed} = 45 + 75 = 120 \ \text{km/h}
2

Work out the time taken to meet

The time to close the gap is the total distance divided by the closing speed. Dividing 160 km by 120 km/h gives the travelling time before they meet.

T=160120=43 h=1 h 20 minT = \frac{160}{120} = \frac{4}{3} \ \text{h} = 1 \ \text{h} \ 20 \ \text{min}
3

Add the time to the start time

Both vehicles set off at 11:50 a.m., so adding the 1 hour 20 minutes travelling time gives the clock time at which they meet at C.

11:50 a.m.+1 h 20 min=1:10 p.m.11{:}50 \ \text{a.m.} + 1 \ \text{h} \ 20 \ \text{min} = 1{:}10 \ \text{p.m.}
4

Find the distance from A to C

Point C is exactly as far as the lorry managed to travel in that meeting time. Multiplying the lorry's speed by the 4/3 hours gives the distance AC.

AC=45×43=60 kmAC = 45 \times \frac{4}{3} = 60 \ \text{km}
5

Find when the lorry reaches B

After C the lorry still has the rest of the road to cover, which is 160 km minus the 60 km already done. Dividing that remaining 100 km by 45 km/h gives the extra time, then adding it to the 1:10 p.m. meeting time gives the arrival time at B.

160−60=100 km160 - 60 = 100 \ \text{km}
10045=2 h 1313 min\frac{100}{45} = 2 \ \text{h} \ 13\tfrac{1}{3} \ \text{min}
1:10 p.m.+2 h 1313 min=3:2313 p.m.1{:}10 \ \text{p.m.} + 2 \ \text{h} \ 13\tfrac{1}{3} \ \text{min} = 3{:}23\tfrac{1}{3} \ \text{p.m.}
6

Find the time the car has for the return trip

The car reached C at 1:10 p.m. and rested for 1 hour 40 minutes, so it left C at 2:50 p.m. It must reach A at the same moment the lorry reaches B, so the travelling time is the gap between 2:50 p.m. and 3:23 1/3 p.m.

1:10 p.m.+1 h 40 min=2:50 p.m.1{:}10 \ \text{p.m.} + 1 \ \text{h} \ 40 \ \text{min} = 2{:}50 \ \text{p.m.}
3:2313−2:50=3313 min=59 h3{:}23\tfrac{1}{3} - 2{:}50 = 33\tfrac{1}{3} \ \text{min} = \frac{5}{9} \ \text{h}
7

Calculate the car's average speed from C to A

The car must cover the same 60 km from C back to A in that 5/9 hour. Dividing the distance by the time gives the required average speed.

Speed=6059=60×95=108 km/h\text{Speed} = \frac{60}{\frac{5}{9}} = 60 \times \frac{9}{5} = 108 \ \text{km/h}

Final Result

The two vehicles meet at 1:10 p.m., town C is 60 km from town A, and the car must travel from C back to A at an average speed of 108 km/h.

Why this method works

The whole problem rests on the idea of relative speed. When two objects move directly towards each other, each metre the first covers plus each metre the second covers both shrink the gap, so the gap closes at the sum of the two speeds — that is why we add 45 and 75 rather than doing anything more complicated. Once the meeting is fixed in time and place, the rest is bookkeeping with the basic relationships distance = speed × time and speed = distance ÷ time. The car's return speed is forced by a timing constraint: it must arrive exactly when the lorry does, so we first pin down that arrival clock time, subtract the rest period and the departure time to find how long the car actually has, and then divide the fixed 60 km by that time.

✓

Check: at 120 km/h closing speed the vehicles do cover 160 km in 4/3 h, and 108 km/h over 5/9 h gives 108 × 5/9 = 60 km, matching distance CA.