Here's the question: An 18 y/o runner can complete a 10.0km course with an avg speed of 4.39 m/s. A 50 y/o runner can cover the same course with an avg speed of 4.27 m/s. How much later (in seconds) should the younger runner start in order to finish the course at the same time as the older runner?
2007-01-18 11:50:57 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Physics
10 kilometres is 10,000 metres.
18 year old runs at 4.39 m/s
10,000 divided by 4.39 = 2277.9 seconds
Thus it takes him 2277.9 seconds to complete the course
50 year old runs at 4.27 m/s
10,000 / 4.27 = 2341.9
Thus it takes him 2,341.9 seconds to complete the course
2341.9 - 2277.9 = 64
!8 year old must give 50 year old a 64 second start
2007-01-18 11:58:00 · answer #1 · answered by Anonymous · 0⤊ 0⤋
do 1 divided by each persons ave speed to give u s/m and then times that by 10000(meters in the race) and then work out the difference...
10000(1/4.39)=2277.9
10000(1/4.27)=2341.9
the difference is 64 so your answers one minuete and four seconds
2007-01-18 12:04:31 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Time1 = 10000 m / (4.39 m/s) = 2278 seconds
Time2 = 10000 m / (4.27 m/s) = 2342seconds
Difference = 2342 - 2278 = 64 seconds
64 seconds or 1 minute 4 seconds after
2007-01-18 11:58:48 · answer #3 · answered by Sheen 4 · 0⤊ 0⤋
All that is necessary to solve this is to compute the running times of each runner, and that is simply distance divided by speed. Take the difference in times, and you're done.
2007-01-18 12:00:30 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I think its 64 seconds. Figure out how long it will take each of them to run by dividing the distance (make sure you convert to meters) by the average velocity and then find the difference in the times.
2007-01-18 11:57:38 · answer #5 · answered by adnama 3 · 0⤊ 0⤋
10km= 10000m
x=vt
10000m=(4.39m/s)t
t= 2277.9s
in order for the 50yo to finish the race at the same time as the 18yo then the 50yo should run ahead of the 18yo. That means time for the 50yo to run is 2277.9+t
x=vt
10000= (4.27)(2277.9+t)
10000=9726.6+4.27t
t= 64.028s
The 50yo should run 64.028s ahead of the 18yo
2007-01-18 12:04:54 · answer #6 · answered by 7 · 0⤊ 0⤋
The 50y/o will complete in 2341.92 seconds. In the same time the 18y/o would cover 10281 meters, 281 meters too long. To cover 281 meters would take the 18y/o 31.25 seconds. The 18y/o should wait 31.25 seconds.
2007-01-18 12:02:52 · answer #7 · answered by Anonymous · 0⤊ 0⤋
64.01604 seconds, rounded to 5 decimal places. All you need to do is divide 10,000m by the two m/s times and subtract the smaller from the larger.
2007-01-18 11:57:01 · answer #8 · answered by Taliesin Pen Beirdd 5 · 0⤊ 0⤋
Time = distance divided by time. So i get it as 64 seconds. But i'd double check!
2007-01-18 11:55:20 · answer #9 · answered by Ben G 1 · 0⤊ 0⤋
We want to find the time difference.
divide distance by speed and get time 10km = 10,000 meters.
Old = 10000/4.27
Young = 10000/4.39
Now subtract Young from Old (old-young) and you'll get the time difference.
2007-01-18 11:54:53 · answer #10 · answered by John T 6 · 0⤊ 0⤋