A school minibus takes students to a sports venue. It travels \(9\) km through a town, taking \(15\) minutes, then continues \(16\) km on a faster road, tak...
Question 1 Report
A school minibus takes students to a sports venue. It travels \(9\) km through a town, taking \(15\) minutes, then continues \(16\) km on a faster road, taking a further \(20\) minutes.
- Work out the minibus\'s speed through the town, in km/h. (1)
- Work out the minibus\'s speed on the faster road, in km/h. (1)
- Work out the average speed for the whole journey, in km/h, correct to \(3\) significant figures, and state which stage was faster. (2)
Answer Details
(a) Speed is distance divided by time. Through the town, the minibus travels \(9\) km in \(15\) minutes, which is \(\dfrac{15}{60}=0.25\) hours, so its speed is \(9\div0.25=36\) km/h. [1 mark]
(b) On the faster road, it travels \(16\) km in \(20\) minutes, which is \(\dfrac{20}{60}=\dfrac{1}{3}\) hours, so its speed is \(16\div\dfrac{1}{3}=48\) km/h. [1 mark]
(c) Average speed for the whole journey uses the total distance and total time, not the average of the two individual speeds: total distance \(=9+16=25\) km, total time \(=15+20=35\) minutes \(=\dfrac{35}{60}=\dfrac{7}{12}\) hours, so average speed \(=25\div\dfrac{7}{12}=42.9\) km/h, correct to 3 significant figures. Since \(48 > 36\) km/h, the faster road stage was the faster of the two. [2 marks]
A common error is averaging \(36\) and \(48\) directly to get \(42\) km/h; because the two stages take different amounts of time, the correct average must be weighted by total distance over total time, which is why it comes out slightly higher, at \(42.9\) km/h.
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