A train starts from rest with constant acceleration a for time T. It then decelerates uniformly with magnitude b until it stops. Its average speed over the complete journey is:
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A train starts from rest with constant acceleration a for time T. It then decelerates uniformly with magnitude b until it stops. Its average speed over the complete journey is:

G1 Physics · Accepted Answer

Maximum speed after the acceleration stage: v = aT. Acceleration-stage distance: s₁ = aT²/2. Deceleration time: t₂ = v/b = aT/b. Deceleration distance: s₂ = v²/(2b) = a²T²/(2b). Total distance: s = aT²(a+b)/(2b). Total time: t = T + aT/b = T(a+b)/b. Therefore average speed = s/t = aT/2.