A man runs a distance of 9km at a constant speed for the first 4 km and then 2 km\h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4 km is
Let the average speed for the first 4 km = x km/h.
Hence, the last 5 km, speed = (x + 2) km/h
Recall: \(Time = \frac{Distance}{Speed}\)
Total time = 1 hour.
\(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\)
\(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\)
\(9x + 8 = x^{2} + 2x\)
\(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\)
\(x^{2} - 8x + x - 8 = 0\)
\(x(x - 8) + 1(x - 8) = 0\)
\(x = -1; x = 8\)
Since speed cannot be negative, x = 8km/h.
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