two cars traveled in the same direction along a straight highway, one at a constant speed of 55mi/h and the other at 70mi/h. how far must the faster car travel before it has a 15-min lead on the slower car?
2006-08-26 18:23:21 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The faster car traveled ~65 miles before it has a 15-min lead on the slower car. See below:
the distance travel when the faster car has 15 min lead:
d' = 15 min. X 55 mi/h = 1/4 hr X 55 mi/hr = 13.75 mi.
distance traveled for slower car:
d = v X t = 55 mi X t
so:
t = d / 55 mi
distance traveled for faster car:
D = V X t = 70 X t
distance traveled for the faster car when it has 15 min lead:
D = d + d' = 70 X t
d + 13.75 = 70 X t = 70 X d / 55
d + 13.75 = 1.27d
.27d = 13.75
d = 50.9
D = d + d' = 50.9 + 13.75 = 64.65 ~65 miles
2006-08-26 18:47:34 · answer #1 · answered by Bet 2 · 0⤊ 0⤋
When both cars have travelled for 1 hour, the slow car will have reach 55 miles and the fast car will have reached 70 miles, putting the faster car 15 miles ahead of the slower car.
Oops--sorry, after I posted this I re-read your question and see you asked when the fast car would have 15 MINUTE lead, not a 15 MILE lead. Sorry. That one will take more work than I'm interested in doing at the moment. :-P
2006-08-27 01:27:23 · answer #2 · answered by Ms.ADJ 2 · 0⤊ 0⤋
First, a 15 min lead is the same as 15/60 hour lead or 0.25 hour lead.
0.25 hr x 55 mi/hr = 3.75 mi
To get this 3.75 mile lead, both cars have to travel for T hours:
3.75 mi = 70 mi/hr x T hr - 55 mi/hr x T hr
3.75 = (70-55)T
T = 3.75/(70-55) = 0.91667 hr
The distance that the faster car would have traveled in T hours is:
70 mi/hr x 0.91667 hr = 64.1667 mi
2006-08-27 01:48:59 · answer #3 · answered by Kitiany 5 · 0⤊ 0⤋
let's give it a try.
the slower car goes 55mph, or .9166 miles per minute. therefore a 15 minute lead is .9166x15=13.75 miles between the cars.
so how far do these two travel in order to put 13.75 miles between them?
70mph travels 1.1666 miles per minute
55mph travels 0.9166 miles per minute
let "x" be the time traveled by each car
then time traveled by 70mph minus time traveled by 55mph = 13.75, which is the 15 minute lead by the faster car.
so: 1.1666(x) - .9166(x) = 13.75
.25(x) = 13.75
x=55 minutes
at the rates listed above, each car therefore traveled:
1.1666x55=64.16 miles
.9166x55=50.41 miles
64.16 - 50.41 = 13.75 miles, which is the 15 minute lead.
i hope.
2006-08-27 01:45:30 · answer #4 · answered by The Beast 6 · 0⤊ 0⤋
the car going 70will drive64.4 miles,stop- 15 min. later, the car going 55 will be there
2006-08-27 01:50:18 · answer #5 · answered by Anonymous · 0⤊ 0⤋
I'm not the best at math, but I think you need to use the algebraic formula of d=rt... meaning distance = rate multiplied by time... sorry I can't be of more help!
~God bless
2006-08-27 01:44:29 · answer #6 · answered by Girl 4 God 3 · 0⤊ 0⤋
55/4=13.75
anser is
13.75miles
2006-08-27 06:05:08 · answer #7 · answered by Navdeep B 3 · 0⤊ 0⤋
by most smart peoples definitions, this is not a good math q, let alone really good math q, its really simple and boring
2006-08-27 02:38:32 · answer #8 · answered by hanumistee 7 · 2⤊ 0⤋
i know the answer but it takes time to type it here... maybe you can email me ur email and i'll type everything up....!
2006-08-27 01:47:27 · answer #9 · answered by RinCo 2 · 0⤊ 1⤋
pshhh, basic algebra, i'm so smart, im not gunna anser that
2006-08-27 01:29:42 · answer #10 · answered by Pie Man 5 · 0⤊ 4⤋