At 7:00 a.m. Joe starts jogging at 6 mi/h. At 7:10 a.m. Ken starts off after him. How fast must Ken run to overtake Joe at 7:30 a.m.?
I did it, but it did not match my choices. Can someone please help me here. I obviously did it wrong. :(
2006-11-20 09:48:38 · 6 answers · asked by Lizzie 5 in Science & Mathematics ➔ Mathematics
Joes' running time: 7:30 - 7:00 = 30 min.
Distance = Speed x time.
Distance = 6 mi/h x ½ h
Distance = 3 mi
Kens' running time: 7:30 - 7:10 = 20 min. (that is, a third of an hour).
Distance = Speed x time.
Speed = Distance / time.
Speed = 3 mi / (1/3)
Speed = 9 mi/h
2006-11-20 10:05:12 · answer #1 · answered by Brenmore 5 · 0⤊ 0⤋
6 miles/h,7 to 7;30 he went 3 miles,ken start 7;10 which means he have only 20 minutes to over take joe for the distance of 3miles . 7;30 -7;20=20 there for 20 /3 =6.67miles per hr ken has to run
2006-11-20 10:28:20 · answer #2 · answered by tefya 1 · 0⤊ 0⤋
Ken would have to run at a 9 mi/hr pace.
set up a proportion
Joe:
6mi/1hr * 1hr/2 = 3mi
Ken:
xmi/1hr * 1hr/3 = 3mi
solve for x and you get 9!
2006-11-20 09:55:02 · answer #3 · answered by bartathalon 3 · 0⤊ 0⤋
9 mph
Ken will cover 3 miles in the half hour. To do the same 3 miles in 20 minutes his speed needs to be 9mph
2006-11-20 09:51:44 · answer #4 · answered by Anonymous · 0⤊ 1⤋
It's 9 mph.
Oh and is that guys picture creepy(the one above)?
Good luck!
2006-11-20 09:55:16 · answer #5 · answered by chromecranium 3 · 0⤊ 0⤋
i am sorry you got it wrong, but i cant tell you the answer, so i will help. Try multiplying the numbers instead of dividing, that is what i did.
2006-11-20 09:53:23 · answer #6 · answered by minnie_mouse♥ 1 · 0⤊ 0⤋