Question

If a body travels 10 m distance at a speed of 10 m/s and then next 10 m distance at a speed of 20 m/s in the same direction, what is its average speed in m/s?

• A. 13
• B. 5
• C. \( \frac{40}{3} \)
• D. \( \frac{50}{3} \)

Hint:

The average speed is the total distance divided by the total time. Use this formula for cases where different speeds are involved over equal distances.

Answer 1

The Correct Option is C

Step 1: Since the two distances are equal, each contributes half of the total distance, so the average speed is the weighted harmonic mean with equal weights: \( \dfrac{1}{v_{avg}} = \dfrac{1}{2}\cdot\dfrac{1}{v_1} + \dfrac{1}{2}\cdot\dfrac{1}{v_2} \).
Step 2: Substitute \( v_1 = 10 \) and \( v_2 = 20 \): \( \dfrac{1}{v_{avg}} = \dfrac{1}{2}\left(\dfrac{1}{10}\right) + \dfrac{1}{2}\left(\dfrac{1}{20}\right) = 0.05 + 0.025 = 0.075 \).
Step 3: Invert to get the average speed: \( v_{avg} = \dfrac{1}{0.075} = \dfrac{40}{3} \, \text{m/s} \).
\[ \boxed{v_{avg} = \frac{40}{3} \, \text{m/s}} \]

Answer 2

A quicker conceptual check uses how much time is actually spent at each speed. The body takes \( t_1 = \frac{10}{10} = 1 \, \text{s} \) at the slower speed and \( t_2 = \frac{10}{20} = 0.5 \, \text{s} \) at the faster speed. Because it spends twice as much time at 10 m/s as at 20 m/s, the average speed must be weighted more heavily toward 10 m/s than toward 20 m/s — it should sit well below the simple average of 15 m/s. Let's check each option against this reasoning.

1. 13: This is below 15 as expected, but it is only an approximation of the true time-weighted value, not the exact result of the calculation.
2. 5: This is far below even the slower speed of 10 m/s, which is impossible since the average of two speeds can never be lower than the smaller of the two.
3. \( \frac{40}{3} \): Using the actual time-weighted average, \( v_{avg} = \frac{v_1 t_1 + v_2 t_2}{t_1+t_2} = \frac{10(1)+20(0.5)}{1+0.5} = \frac{20}{1.5} = \frac{40}{3} \, \text{m/s} \), which is below 15 m/s as expected and matches the reasoning exactly.
4. \( \frac{50}{3} \): At about 16.67 m/s, this is above 15 m/s, contradicting the fact that more time is spent at the slower speed — this option would only make sense if more time were spent at 20 m/s, which is not the case here.

The time-weighted calculation confirms the average speed is \( \frac{40}{3} \) m/s, consistent with it being pulled toward the slower speed.

Therefore, the correct answer is \( \frac{40}{3} \) m/s.