Speed at the midpoint of a uniformly accelerated journey

The key is to distinguish the midpoint of the distance from the midpoint of the time. They are generally not the same event.

Problem

A particle covers a distance in one direction with constant acceleration. Its average velocity over the complete journey is V_av. Find the possible range of the magnitude of its velocity at the midpoint of the path.

uinitial speed

vfinal speed

v_mspeed at distance s/2

V_avaverage velocity

First: which “midpoint”?

Do not average u and v at the spatial midpoint.

For constant acceleration, (u + v)/2 is the velocity at the midpoint of the time interval. This problem asks for the velocity after half the distance has been covered.

Build the two required relations

Use the given average velocity

Because acceleration is constant and motion remains in one direction,

V_av = (u + v)/2 ⇒ u + v = 2V_av.
Use velocity as a function of distance

Let the complete distance be s. At the end of the path,

v² = u² + 2as.

At the spatial midpoint, the displacement is s/2, so

v_m² = u² + 2a(s/2) = u² + as.

Eliminating as gives the central result:

v_m² = (u² + v²)/2.

See the range graphically

Choose the initial speed u as a fraction of V_av

u = 0.35 V_av

Initial speed0.35 V_av

Midpoint speed1.19 V_av

Final speed1.65 V_av

Speed squared varies linearly with distance The central dot is at half the path, not half the time.

Midpoint speed versus initial speed The curve reaches its highest point at u = 0.

Why does the upper bound occur at u = 0?

Substitute v = 2V_av − u into the midpoint relation:

v_m² = [u² + (2V_av − u)²]/2 = V_av² + (V_av − u)².

We may first consider the accelerating orientation, so 0 ≤ u ≤ V_av. The first term above is fixed. Therefore, maximizing v_m is exactly the same as maximizing the distance of u from V_av. Within the allowed interval, that distance is greatest at the endpoint u = 0.

Upper bound: greatest possible inequality

Set u = 0. Then v = 2V_av, and

v_m,max = √2 V_av.

The particle starts from rest and gains speed throughout the journey.

Lower bound: no inequality at all

Set u = v = V_av. Then the acceleration is zero, and

v_m,min = V_av.

The particle moves with constant velocity throughout the journey.

Final answer

The magnitude of the velocity at the midpoint of the path can lie in the range

V_av ≤ |v_m| ≤ √2 V_av.

The lower equality corresponds to uniform velocity; the upper equality corresponds to starting from rest (or, in the time-reversed motion, ending at rest).

Endpoint convention. If “unidirectional” is interpreted as requiring a strictly positive speed even at the endpoints, then u = 0 is not included. In that stricter interpretation, √2 V_av is approached as u → 0⁺ but is not attained. Standard textbook usage allows the particle to be momentarily at rest at an endpoint.