Raymond B. answered 10/03/20
Math, microeconomics or criminal justice
To collide they must go different speeds. Let their speeds be x, y, and z for the plows leaving a 12, 1 and 2pm
slowest is the one leaving at noon
fastest is last to leave, at 2pm, if they are all 3 to collide at the same time
x<y<z
distance traveled is the same for each plow
d=rt where d=distance, r = rate of speed and t = time
for the 1st plow, d=xt
for the 2nd plow d =y(t-1)
for the 3rd d = z(t-2)
xt= y(t-1) = z(t-2)
xt=yt-y
yt-xt=y
t(y-x)=y
t=y/(y-x)
12+t = time of the collision = 12 + y/(y-x)
xt=zt-2z
zt-xt=2z
t(z-x) = 2z
t = 2z/(z-x)
12 +t = time of the collision = 12 + 2z/(z-x)
y(t-1) = z(t-2)
yt -y = zt -2z
zt - yt = y-2z
t(z-y) = (y-2z)
t = (y-2z)/(z-y)
12 + t = time of the collision = 12 + (y-2z)/(z-y)
If you know any two of the plows' speeds you can calculate the time of the collision
Odds are snow started before the first plow left, sometime before noon, depending on fast it came down. The heavier the snow, the closer to noon was the first snow fall.
but maybe there was more information not given in the problem?