In 2014 Meb Keflezighi won the famous Boston Marathon with a time of 2:08:37. Be

Question

In 2014 Meb Keflezighi won the famous Boston Marathon with a time of 2:08:37. Below you will find a table representing his split times during the race. Using at least a second order accurate finite difference approximation, calculate Meb's average speed (in mph) at each stage of the race (you can ignore the " half" and " finish" point since they are not equally spaced compared to the rest of the points). Note that you will have to use different finite difference approximations for the end points and inner points.

Explanation / Answer

i)at 5 km the time taken is 15 min 9 sec which is 15*60 + 9 = 909 sec = 0.2525 hr

so speed = 5/ 909 km/sec = 5000/0.2525= 19801 m/hr

finite difference value at 5 km = (19801 - 0) /5 = 3960.2 m/hr

average speed at 5km = 3960.2 m/hr

ii)at 10 km the time taken is 30 min 29 sec which is 30*60 + 29 = 1829 sec = 0.51 hr

so speed at 10km = 10/ 1829 km/sec = 10000/0.51 = 19607 m/hr

finite difference value at 10 km = (19607 - 19801) /5 = 38.8 m/hr

average speed = (3960.2+38.8)/2 = 3999/2 = 1999.5m/hr

iii)at 15 km the time taken is 45 min 47 sec which is 45*60 + 47 = 2747 sec = 0.76 hr

so speed = 15/ 2747 km/sec = 15000/0.76= 19737 m/hr

finite difference value at 15 km = (19737 - 19607) /5 = 26 m/hr

average speed at 15km = (3960.2+ 38.8+26)/3 = 1342 m/hr

iv)at 20 km the time taken is 1 hr 1 min 5 sec which is 3600 + 60 + 5 = 3665 sec = 1.02 hr

so speed = 20/ 3665 km/sec = 20000/1.02= 19607 m/hr

finite difference value at 20 km = (19737 - 19607) /5 = 26 m/hr

average speed at 20 km = (3960.2+ 38.8+26+26)/4 = 1013 m/hr

v)at 25 km the time taken is 1 Hr 16 min which is 3600 + 16*60 = 3600+960 sec = 4560 sec = 1.3 Hr

so speed = 25/ 4560 km/sec = 25000/1.3 = 19231 m/hr

finite difference value at 25 km = (19231 - 19607) /5 = 75 m/hr

average speed at 15km = (3960.2+ 38.8+26+26+75)/5 = 825.2 m/hr

vi)at 30 km the time taken is 1 hr 31 min 10 sec which is 3600 + 31*60+10 = 5470 sec = 1.52 hr

so speed = 30/ 5470 km/sec = 30000/1.52 = 19737 m/hr

finite difference value at 30 km = (19737 - 19231) /5 = 101 m/hr

average speed at 30km = (3960.2+ 38.8+26+26+75+101)/6 = 705 m/hr

vii)at 35 km the time taken is 1 Hr 46 min 37 sec which is 3600+46*60 + 37 = 6397 sec = 1.8 hr

so speed = 35/ 6397 km/sec = 35000/1.8 = 19444 m/hr

finite difference value at 35 km = (19737 - 19444) /5 = 59 m/hr

average speed at 35km = (3960.2+ 38.8+26+26+75+101+59)/7 = 4286/7 m/hr= 612.3 m/hr

viii) at 40 km the time taken is 2 Hr 1 min 49 sec which is 7200+60 + 49 = 7309 sec = 2.03 hr

so speed = 40/ 7309 km/sec = 40000/2.03 = 19704 m/hr

finite difference value at 40 km = (19704 - 19444) /5 = 52 m/hr

average speed at 40km = (3960.2+ 38.8+26+26+75+101+59+52)/8  = 4286/7 m/hr= 542.3 m/hr

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