In 1954 the English runner Roger Bannister broke the four minute barrier for the mile with a time of 3:59.4 s (3 min and 59.4 sec). In 1999 the Moroccan runner Hicham el-Guerrouj set a record of 3:43.13 s for the mile. If these two runners had run in the same race, each running the entire race at the average speed that earned him a place in the record books, el-Guerrouj would have won. By how many meters?
2007-06-05 05:37:12 · 4 answers · asked by M&M 2 in Science & Mathematics ➔ Physics
multiply Bannister average speed by 223.13 seconds to find the distance he has traveled in the time it takes el-Guerrouj to finish the race. Subtract that distance from 1 mile (5280 ft), then convert feet to meters.
2007-06-05 05:44:39 · answer #1 · answered by SteveA8 6 · 1⤊ 0⤋
1 mile = 1.61km = 1610m
time req for Roger Bannister
= (3 x 60 + 59.4)s
= 239.4s
avg speed of Roger Bannister
= (1610m / 239.4s) = 6.7251462 m/s
time req for el-Guerrouj
= (3 x 60 + 43.13)s
= 223.13s
in 223.13s Roger Bannister covered
(6.7251462 x 223.13)m
= 1500.58m
el-Guerrouj won by (1610 - 1500.58)m
= 109.42m
2007-06-05 12:52:49 · answer #2 · answered by src_virus 2 · 2⤊ 0⤋
When el-Guerrouj wins, 3:43.13 minutes, that is, 223.13 seconds have passed by.
In that time, Bannister would have run
(223.13 / 239.4) = 0.932 miles
Therefore, el-Guerrouj wins by
1 - 0.932 = 0.068 miles,
which is
0.068 * 1 604 = 109 meters.
2007-06-05 12:49:00 · answer #3 · answered by dutch_prof 4 · 2⤊ 0⤋
Average speed of Bannister = 5280 / 239.4 = 22.055 ft/sec (we will convert it to metres finally)
Average speed of Guerrpuj = 5280/223.13 = 23.663 ft/sec
When Hicham reached the finsih line, he will be ahead of Roger by (239.4 - 223.13) X (23.663) ft = 384.997 ft and this we can convert to meteres = 384.997 X 12 X 2.54 / 100 = 117.347 m
2007-06-05 12:46:10 · answer #4 · answered by Swamy 7 · 0⤊ 1⤋