Two components in a stated learner model
A fictional theory worksheet explicitly defines reaction distance as(v/10)times 3 metres and ordinary braking distance as(v/10)squared metres, wherev is the supplied speed in kilometres per hour. It asks for the approximate total at 50 and 100 km/h. The task assumes the same model conditions and makes no claim that an actual vehicle will always stop within those distances. A learner doubles every component when the speed doubles and another applies an emergency-braking adjustment to the reaction part as well. No real surface, tire, gradient, brake system or personal reaction measurement is provided. A correct answer shows the two calculations separately and identifies which changes with the square of speed, rather than treating the result as permission to travel at a particular speed on an unseen road.
1. Compute the two total distances.
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At 50 km/h, reaction is 15 m and braking 25 m, totaling 40 m. At 100 km/h, reaction is 30 m and braking 100 m, totaling 130 m under the given model.
2. Does doubling speed double total distance?
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No. Reaction doubles in this model but braking quadruples. The total changes from 40 to 130 m, and actual stopping also depends on conditions absent from the worksheet.