at 7 am joe starts jogging at 6mi/h at 7:10am ken starts off after him. how fast must ken run in order to overtake him at 7:30 am?
2007-06-11 17:32:18 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
well he should hop in his car and he will catch up with him in no time...
2007-06-11 17:37:49 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Joe begins at 7:00 am and runs at 6mph. Therefore, using d=rt, plug in 6 for d and .5 (half an hour) for t. So, at 7:30 he has ran 3 miles.
So, you then know Ken needs to run 3 miles in 20 minutes. Again use d = rt, 3 for d, 1/3 for t (20 minutes is 1/3 of an hour). Divide and you get 9 mph.
2007-06-12 00:41:46 · answer #2 · answered by latus rectum 1 · 0⤊ 0⤋
9 mph
he has to run the 3 miles joe ran in 30 mins. in 20 mins
2007-06-12 00:40:41 · answer #3 · answered by Awaitng hurricane season 3 · 0⤊ 0⤋
time ken runs, 1/3 hr
time joe runs, 1/2 hr
distance = speed * time, so
6(1/2) = r(1/3)
3 = r/3
r = 9 mph
2007-06-12 00:41:39 · answer #4 · answered by Philo 7 · 0⤊ 0⤋
9 mi/h
2007-06-12 00:39:50 · answer #5 · answered by Coventry_ 1 · 0⤊ 0⤋
do your own homework!!!
why wasn't ya around when i was in school!!
2007-06-12 00:39:45 · answer #6 · answered by ? 4 · 0⤊ 0⤋