John ran a race at his school. The course was measured at 450m correct to 2sf and his time was given at 62 econds to the nearest second. Calculate the difference between his maximum and minimum possible average speed. Round you answer to 3sf.

Upper bound for distance: 455mLower bound for distance: 445mUpper bound for time: 62.5sLower bound for time: 61.5sSpeed = Distance/timeMaximum possible average speed would be the longest distance divided by the fastest time i.e. upper bound for distance, divided by lower bound for time:455/61.5 = 910/123 m/sMinimum possible average speed would be the shortest distance divided by the slowest time i.e. lower bound for distance, divided by upper bound for time:445/62.5 = 178/25 m/sThe difference is therefore: 910/123 - 178/25 = 0.2783739837 m/s = 0.278 m/s to 3s.f.

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Answered by Gideon E. Maths tutor

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