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.

GE
Answered by Gideon E. Maths tutor

4143 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

If two linear equations, y = x + 4 and y = 2x + c, intersect at x = 1, find c.


x = 0.436363636... . Prove algebraically that x can be written as 24/55.


3 teas and 2 coffees have a total cost of £7.80. 5 teas and 4 coffees have a total cost of £14.20. Work out the cost of one tea and the cost of one coffee.


Could you explain ratios to me?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning