1. Terri begins walking east at 2 miles per hour at 1 p.m. If cindy leaves from the same point 30 minutes later walking wast at 3 miles per hour, when will she catch Terri?... Show Work and Explain Please..
2006-12-12 14:18:00 · 3 answers · asked by SaY wHaT >? 2 in Science & Mathematics ➔ Mathematics
This question could be very interesting if you meant "west" instead of "wast", not so interesting if you meant "fast" instead of "wast"!
2006-12-12 14:21:16 · answer #1 · answered by cfpops 5 · 0⤊ 0⤋
Let t = hours after 1:30. At time t >= 0, Terri's position will be 1 + 2t (she has walked 1 mile by 1:30 and is continuing at 2 miles per hour) and Cindy's position will be 3t. Cindy will catch Terri when these are equal, i.e. when 1 + 2t = 3t, i.e. at t = 1 which corresponds to 2:30 pm.
If you did mean that Cindy is walking west, it depends on the latitude and how fast and far they can each swim - most likely they will both drown!
2006-12-12 22:27:59 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
Well - Cindy gave terri a 1 Mile head start, then we calulate time from the time that cindy left.
Distance = rate * time
Terri's Distance = 2 mi/hr * X hours + 1 mile
Cindy's Distance = 3mi/hr * X hours.
(yes X is the same, because once the 1 mile head start is given, they are on the same time)
D = r * t
2 mi/hr * x + 1 (the headstart)= 3mi/hr * x
2x +1 = 3
2x = 2
x = 1hour
So after 1 hour they will both have travelled the same distance.
But since Cindy is heading West, she will not catch Terri unless they both can walk on water.
2006-12-12 22:25:11 · answer #3 · answered by holst22 1 · 1⤊ 0⤋