Q.A vehicle travels half the distance $L$ with speed $V_1$ and the other half with speed $V_2$, then its average speed is

Concept: Average speed is total distance divided by total time — not the arithmetic mean of the speeds.

• Total distance = $L$.
• Time for first half: $t_1 = \frac{L/2}{V_1} = \frac{L}{2V_1}$.
• Time for second half: $t_2 = \frac{L/2}{V_2} = \frac{L}{2V_2}$.
• Total time = $t_1 + t_2 = \frac{L}{2}\left(\frac{1}{V_1} + \frac{1}{V_2}\right) = \frac{L(V_1 + V_2)}{2V_1 V_2}$. …

The trap here is to reach for the arithmetic mean $\frac{V_1 + V_2}{2}$ — that would be correct only if the vehicle spent equal time at each speed, not equal distance. Since the halves are equal in distance, the time spent at each speed is different, and the average speed must be computed directly.

Why instantaneous velocity matters here: Average speed is defined as total path length divided by total time. Instantaneous velocity tells us how fast the vehicle is moving at any moment, but to find the average we need to integrate (or sum) the time intervals. For constant speeds over segments, this reduces to a simple weighted average — weighted by time, not distance.

Let’s work it through.

1. Define the total distance.

   The total distance is $L$. The first half is $\frac{L}{2}$ at speed $V_1$, the second half is $\frac{L}{2}$ at speed $V_2$.

2. Find the time for each half.

   Time = distance / speed.

3. Total time.

4. Average speed.

5. Simplify. Combine the fractions in the denominator: …

Concept: Scaling Away the Irrelevant Variable (WLOG $L=1$)

Method: Nondimensionalize First (set $L=1$ by scale invariance, then generalize back)

The answer must be a pure combination of $V_1$ and $V_2$ only — the specific distance $L$ cannot appear in it, because "average speed" for a trip covering half of $L$ at each speed is a ratio that stays the same no matter how long the road actually is. This method exploits that scale-invariance directly, instead of carrying the symbol $L$ through the whole calculation only to watch it cancel at the end.

Steps

1. Justify why $L$ must drop out. If the road were twice as long, both half-distances double, both times $t_1,t_2$ double proportionally, and the ratio (total distance)/(total time) — average speed — is unchanged. So average speed here is a function of $V_1,V_2$ alone; $L$ is a scale factor that cannot survive into the final formula.

2. Since $L$'s actual value is irrelevant, set $L=1$ without loss of generality — this is purely a computational convenience, not an approximation.

3. Compute the times for each half, with $L=1$:

4. Total time:

5. Average speed $=\dfrac{\text{total distance}}{\text{total time}} = \dfrac{1}{T}$ (since distance was set to $1$): $v_{\text{avg}} = \frac{1}{\dfrac{V_1+V_2}{2V_1V_2}} = \frac{2V_1V_2}{V_1+V_2}$ …