Repeat the question if the walker walks at the rate of 6mph for 3hrs and then at 7mph
2006-10-06 09:55:43 · 5 answers · asked by primo026 1 in Science & Mathematics ➔ Mathematics
Using distance=rate*time, we know that the person with the head start traveled 3*7=21 miles.
His position, in miles, is 21+7t, where t is in hours.
The other person's position is 17t, so he overtakes the first person when
17t=21+7t, so 10t=21, meaning t=21/10. Therefore, it takes 2 hours and 6 minutes to overtake him.
If the walker walks at 6mph for 3 hrs, then his position is 18+7t, so overtaking occurs when 17t=18+7t, or t=18/10, 1 hour and 48 minutes.
2006-10-06 10:01:50 · answer #1 · answered by James L 5 · 1⤊ 0⤋
Number One
1 hr = Walker 6 miles
Jogger 0 miles
2 hr = Walker 12 miles
Jogger 0 miles
3 hr = Walker 18 miles
Jogger 0 miles
4 hr = Walker 24 miles
Jogger 17 miles
5 hr = Walker 30 miles
Jogger 32 miles
It would take the jogger 2 hrs (5 - the 3 hr head start) to overtake the walker.
Number Two
1 hr = Walker 7 miles
Jogger 0 miles
2 hr = Walker 14 miles
Jogger 0 miles
3 hr = Walker 21 miles
Jogger 0 miles
4 hr = Walker 28 miles
Jogger 17 miles
5 hr = Walker 35 miles
Jogger 32 miles
6 hr = Walker 42 miles
Jogger 49 miles
It would take the jogger 3 hrs (6 - the 3 hr head start) to overtake the walker.
2006-10-06 17:11:59 · answer #2 · answered by NoSuchThing 2 · 0⤊ 0⤋
the fast walker travels 17 miles in 1 hr (which is just ridiculous anyway) but here goes
the other dude travels at 7 miles an hour with a three hour head start gathering 21 miles.
the fast dude is 4 miles short of the slow dude.
in order to pass this guy it would take him the one hour that he does the 17 miles in and a further:
0.235 hrs
this is how i got the figure above:
in 1 hour he does 17 miles 1hr = 17 miles
he has 4miles left to do so if he's travelling at the same speed:
his 4 miles will take (1hour/ 17 miles) * 4 miles = 0.235 hrs
but he needs to pass the slower dude so lets round up and say he takes 1.24hours to pass the slow guy.
2006-10-06 17:13:00 · answer #3 · answered by Al Cho 1 · 0⤊ 0⤋
The runner is closing at 17-7=10mi/hr
the walker started 7mi/hr*3hrs=21mi ahead.
after 21mi/10mi/hr=2.1hrs=2hrs 6min, the runner will catch up.
If you can find a jogger who can average 17mph (averaging about 3:30 per mile) please tell the USOC.
I personally don't know anybody who can walk at a better than 9min/mi pace.
2006-10-06 17:02:46 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
d = 17t = 7(t+3), t in hrs
17t = 7t + 6*3
2006-10-06 17:03:15 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋