John ran a 450m race (2sf) in a time of 62 seconds (nearest second). Calculate the difference between his maximum and minimum average speed. (3sf)

S=D/T Max= 455(upper bound)/61.5(lower bound)= 910/123 m/s Min= 445(lower bound)/62.5 (upper bound) = 178/25 m/s Max - Min= 856/3075 m/s

CR
Answered by Christopher R. Maths tutor

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