at 8:00 a.m. the smiths left a campground, driving at 48 mi/h. At 8.20 a.m. the Garcias left the same campground and followed the smae route, driving at 60 mi/h. At what time did the Garcia family overtake the Smiths?
I would grealty appericte any help. When solving can you please show the process to get the answer. I want see how it is done, so I can know for my other amth problems.
~♥*~
2006-11-27 16:24:50 · 4 answers · asked by Lizzie 5 in Science & Mathematics ➔ Mathematics
I know it has to do with Rate-Time-Distance
2006-11-27 16:25:31 · update #1
I think the easiest to look at this one is:
At 48mph, the Smiths cover 4 miles in 5 minutes.
At 60mph, the Garcias cover 5 miles in 5 minutes.
So, every 5 minutes, the Garcias get 1 mile closer to the Smiths.
The Smiths started 20 minutes before the Garcias, so they were 16 miles down the road when the Garcias started. Therefore, the Garcias needs 16 * 5 = 80 minutes to catch the Smiths.
2006-11-27 16:34:50 · answer #1 · answered by Anonymous · 0⤊ 0⤋
the headstart that the smiths have=20/60*48
=16 miles
relative speed of the garcias=60-48
=12mph
time taken for garcias to overake the smiths=16/12
=16/12*60 min
=1 hr and 20 minutes
they will catch up at 9.40 a.m.
2006-11-28 00:29:41 · answer #2 · answered by raj 7 · 0⤊ 0⤋
in the 1st 20 min =1/3 hr, the smiths travelled 48/3=16 miles
the Garcias, friving 12mph faster closed the 16 miles in
16mi/12mi/hr=4/3 hrs=1 hr 20 min
they caught up at 8:20+1:20=8:40AM
thet were 60*4/3=80 miles from the campground at that time.
2006-11-28 00:28:57 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
the first one will take
16 mi
when the second begins
The length of the distance for them is same
Hence
8.20-8.00=20 minutes=20/60 hour
48*20/60=16 mi
16+48*t=60*t
60t-48t=16
12t=16
t=16/12 hours= 16/12 * 60 minutes=80 minutes
8:20+80 minutes=9:40
2006-11-28 00:44:40 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋