A car climbs a slope of angle α = 10^° at a steady speed v_1…
A car climbs a slope of angle $\alpha = 10^\circ$ at a steady speed $v_1$. Coming down the same slope with the same engine power, its steady speed is $v_2 = 30\ \mathrm{m/s}$. What will its speed be on a level road under the same conditions (same power, same resistance)?
Explanation of Method
Car speed uphill, downhill and on the level.
Step 1: Driving force at constant power
With power $P$, the driving force at speed $v$ is $P/v$. At steady speed it balances the other forces along the road (resistance $F_r$ and the slope component $mg\sin\alpha$):
\[ \text{uphill: } \frac{P}{v_1} = F_r + mg\sin\alpha, \]
\[ \text{downhill: } \frac{P}{v_2} = F_r - mg\sin\alpha, \]
\[ \text{level: } \frac{P}{v} = F_r. \]
Step 2: Eliminate the slope
Adding the first two equations: $\frac{P}{v_1} + \frac{P}{v_2} = 2F_r = \frac{2P}{v}$.
\[v = \frac{2v_1v_2}{v_1 + v_2}\]
The level-road speed is the harmonic mean of the uphill and downhill speeds; neither the angle nor the power is needed.
Step 3: Numbers
The statement does not give $v_1$. With $v_1 = 20\ \mathrm{m/s}$ (the value the original solution assumed): $v = \frac{2 \cdot 20 \cdot 30}{20 + 30}$.
\[v = 24\ \mathrm{m/s} \ \ (\text{for } v_1 = 20\ \mathrm{m/s})\]
Result
\[\boxed{\,v = \frac{2v_1v_2}{v_1 + v_2}\,}\]
\[\boxed{\,v = 24\ \mathrm{m/s} \ \ (\text{for } v_1 = 20\ \mathrm{m/s})\,}\]
Remark
About the original solution. It had the sign of the slope term wrong for the downhill case and then stated $v = \sqrt{v_1v_2}$ without a valid derivation, giving 24.5 m/s. The correct relation is $v = 2v_1v_2/(v_1 + v_2)$, which gives 24 m/s for the assumed $v_1 = 20\ \mathrm{m/s}$. The actual $v_1$ should be taken from the original problem; the formula holds for any value.