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Fig. 1. A model of how stopping ability and total travel distance could constrain modulation of velocity during a start and stop. The hypothetical movements in A and B have identical starting accelerations (astart=slope=1) and total displacements (Stotal=areas under the triangles or trapezoids with a vertex indicated by a circle=3 units), but the stopping acceleration (astop) varies among the cases indicated by the different colors within each panel. (A) Starting acceleration continues up to the instant when stopping begins (indicated by circles). Compared to astop=1, increased stopping ability (green) allows more distance to accelerate to a greater maximal velocity (Vmax), which decreases total travel time. Decreased stopping ability (red) has detrimental effects on Vmax and total travel time. Consequently, the average velocities (Stotal divided by total time) for astop=2 and 0.5 are 115% and 81% of the value when astop=1, respectively. (B) When the total distance provides sufficient time so that a physiologically maximum speed is attained and momentarily sustained, stopping ability will not affect Vmax. However, increased stopping ability decreases total time and hence the average velocities for astop=2 and 0.5 are 107% and 89% of the value when astop=1, respectively. Maintaining a constant velocity in between the starting and stopping accelerations (B) increases total travel time and hence decreases average velocity compared to beginning a stop immediately after the cessation of a starting acceleration (compare A vs B for equal values of astart and astop). If the only objective of starting and stopping is to minimize total travel time (and maximize average speed) for a given distance, then maximal accelerating and decelerating capacities should be used. (C) For a linear increase in velocity followed immediately by a linear decrease in velocity as in A, Vmax=astart[(2Stotal)/(astart+1/astop)]0.5 and hence the upper limit of Vmax is astart0.5(2Stotal)0.5. Axes show arbitrary units.





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