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

3860 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Using the diagram and your knowledge of vectors, show that BCD is a straight line


How do you factorise a quadratic equation


[equ1] 3y − 6x = 3 [equ2] y y x 2 − x + 2 2 = 2


A family go into a shop, they buy three sandwiches and two packets of crisps. It costs them £9. Another family buy five sandwiches and six packets of crisps. It costs them £19. How much does two sandwiches and five packets of crisps cost?


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences