math
posted by tina .
at 7:00am joe starts jogging at 6 mi/h. at 7:10 am ken starts after him. how fast must ken run to overtake joe at 7:30am?
If Joe runs at 6 mi/hr for 30 minutes (7:00 to 7:30), his distance traveled is 6 mi/hr x 0.5 hr = 3 miles.
Ken starts at 7:10 and runs until 7:30 for a time of 20 minutes. For Ken to just catch up with Joe in 20 minutes, the rate  distance/time = 3 mi divided by 1/3 Hr =3 x 3/1 = 9 miles/hr. So to overtake Joe, Ken must run one step faster somewhere in the 3 miles.

this is not a valid question