two cars with average speeds of 40 and 50 km/h respectively moved toward each other at 4:00 am . if the square of the distance traveled by the second car is 250 times the distance traveled by the first car , at what time will they meet?how far has each traveled by then?
2006-10-10 00:43:15 · 7 answers · asked by avenger 2 in Science & Mathematics ➔ Mathematics
Time is T
Car 1 distance: 40*T
Car 2 distance: 50*T
(50*t)**2 = 250*40*T
2500*t**2 = 2500*4*t, so t**2 = 4*t
So there are 2 answers: one answer is T = 0, and they meet at 4:00 AM.
The other nontrivial answer os t = 4, so they meet at 8:00 AM, car 1 travels 160 km, car 2 200.
2006-10-10 00:52:58 · answer #1 · answered by sofarsogood 5 · 1⤊ 0⤋
Say the cars are d distance apart and that the first car goes x distance and the second y.
then: x + y = d -----------(1)
y^2 = 250x ---------------(2)
Since both travel for the same time (= t) before meeting
y = 50t and x = 40t (speed = distance/time)
then y/50 = x/40 ---> x = 4/5 y
from (2)
y^2 = 200y
y = 200 km
x = 160km
t = y/50 = 4hrs=
So they will meet at 8am
2006-10-10 00:59:07 · answer #2 · answered by blind_chameleon 5 · 0⤊ 0⤋
car 1 goes A miles
car 2 goes B miles
b^2=250A
A=(40/50)b=.8B
B=1.25A
B^2=250*1.25B=312.5B
B=312.5 km
A=.8B=.8*312.5=250km
time=B/50=312.5/50=6.25 hrs
2006-10-13 08:27:47 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
Depends on how far the cars were form each other at 4 a.m. Which hasnt been stated in your equation!
2006-10-10 00:49:15 · answer #4 · answered by Anonymous · 0⤊ 1⤋
8am.
160km and 200km.
Let x be the number of hours traveled.
Distance traveled by first car is 40x, by second car 50x.
Equate (50x)^2=250*(40x) and solve.
2006-10-10 00:58:04 · answer #5 · answered by Mark P 5 · 0⤊ 0⤋
I will say that neither has a chance to survive.
2006-10-10 00:47:27 · answer #6 · answered by Dr. J. 6 · 0⤊ 0⤋
Not enough info, how far are they apart?
2006-10-10 00:51:37 · answer #7 · answered by Grev 4 · 0⤊ 1⤋