SOLUTION: A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B; if the accident has occurred
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Question 562910: A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B; if the accident has occurred 50 miles further on, it would have been 2 hours late. Find the distance from A to B and the original velocity of the train? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B;
if the accident has occurred 50 miles further on, it would have been 2 hours late.
Find the distance from A to B and the original velocity of the train?
:
Let s = normal speed of the train
then
.6s = speed after accident
:
d = distance from A to B
then = normal travel time of the train from A to B + 3 = travel time after 50 mi accident + 2 = travel time after 100 mi accident
:
50 mi accident travel time equation + = + 3
multiply by .6s to clear the denominators, results
.6(50) + (d-50) = .6d + .6s(3)
30 + d - 50 = .6d + 1.8s
d - 20 = .6d + 1.8s
d - .6d - 1.8s = 20
.4d - 1.8s = 20
:
100 mi accident travel time equation + = + 2
multiply by .6s, results
.6(100) + (d-100) = .6d + .6s(2)
60 + d - 100 = .6d + 1.2s
d - 40 = .6d + 1.2s
d - .6d - 1.2s = 40
.4d - 1.2s = 40
:
Subtract the 1st equation from the above equaton
.4d - 1.2s = 40
.4d - 1.8s = 20
----------------
+.6s = 20
s =
s = 33 mph is the normal train speed
Find d
.4d - 1.2(33.33) = 40
.4d - 40 = 40
.4d = 40 + 40
.4d = 80
d = 80/.4
d = 200 mi is the distance
;
:
See if the that checks out in the 100 mi accident equation + = + 2
3 + = 6 + 2
3 + 5 = 6 + 2
:
We can say: normal speed = 33 mph; distance is 200 mi
