| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
A NORTH-bound train 88 Yards long meets a SOUTH-bound train also
88 Yards long at 12 o'clock and takes 6 seconds to pass it. At 15
minutes and 3 seconds after 12 it meets another south-bound train
132 Yards long and takes 6 seconds to pass it. When will the second
SOUTH-bound train overtake the first if the rate of the NORTH-bound
train is 35 miles per hour?
Note:
35 miles per hour is = 35*(5280 /3) /(60 * 60) yards per second
= 154/9 yards per second.
Enjoy.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 452.1 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Tue Mar 11 1986 09:30 | 27 | |
1:15, assuming "overtake" means the front of one train passes the
front of the other. I hope they are on different tracks.
To obtain that solution, convert all units to yards, seconds, or
yards per second. Write six functions:
1) position of front of north-bound train:
Vn*t
2) position of back of north-bound train:
Vn*t - 88 yards
3) position of front of first south-bound train:
-Vs*t
4) position of back of first south-bound train:
-Vs*t + 88 yards
5) position of front of second south-bound train:
(Vn*900) - Vx(t-900)
6) position of back of second south-bound train:
(Vn*900) - Vx(t-900) + 132 yards
Time and distance are measured from noon at the place where the first
trains met. North is positive. Set 2 and 4 equal at t=6 and solve
for Vs. Set 2 and 6 equal at t=906 and solve for Vx. Set 3 and
5 equal and solve for t.
-- edp
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