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(1)\begin{aligned}\text { We knowe that speed } & =\frac{\text { Distance }}{\text { Time }} \\\text { Given distance } & =6.2 \mathrm{mi} \\\text { Time } & =32 \text { minutes } \\& =\frac{32}{60} \text { hours } \\\text { Then speed } & =\frac{6.2 \mathrm{mi}}{\frac{32}{60} \text { hours }}=11.625 \mathrm{mi} \text { hours }\end{aligned}By Mean Value theosem, if the average speed is 11.625 milhoure, it means that all speeds between 0 and 11.625 milhe were reached.As the runner's epecd at finish line is 0 , it means he has been running at exactly 11 milhr twice in the race.The first time when accelerating from omilhr to the highest speed andsecond time when going from highest speed too milhr.By the intermediate value theotem, the speed was exactly 11 milhr at least twice.By MVT, all suecds bctween 0 and 11.625 milhr were reached.Becausc the initial and final speed was 0 milhr, the speed of 11 milhr was reached at least turce in ...