A car moves from X to Y with a uniform speed vᵤ and returns to X with a uniform speed v_d. The average speed f

Question

A car moves from X to Y with a uniform speed vᵤ and returns to X with a uniform speed v_d. The average speed for this round trip is:

Accepted Answer

1.  **Recall Definition:** Average speed is defined as the total path length divided by the total time interval .
    $$ v_{avg} = \frac{	ext{Total Path Length}}{	ext{Total Time}} $$
2.  **Define Variables:**
    Let the distance between X and Y be $d$.
    *   For the trip from X to Y: Speed = $v_u$, Distance = $d$, Time $t_1 = \frac{d}{v_u}$.
    *   For the return trip from Y to X: Speed = $v_d$, Distance = $d$, Time $t_2 = \frac{d}{v_d}$.
3.  **Calculate Totals:**
    *   Total Path Length = $d + d = 2d$.
    *   Total Time = $t_1 + t_2 = \frac{d}{v_u} + \frac{d}{v_d} = d \left( \frac{1}{v_u} + \frac{1}{v_d} ight) = d \left( \frac{v_d + v_u}{v_u v_d} ight)$.
4.  **Calculate Average Speed:**
    $$ v_{avg} = \frac{2d}{d \left( \frac{v_d + v_u}{v_u v_d} ight)} $$
    $$ v_{avg} = \frac{2 v_u v_d}{v_u + v_d} $$
    (This is the harmonic mean of the speeds).

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