| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1657.1 | | TAV02::NITSAN | One side will make you larger | Sun Aug 30 1992 09:34 | 9 |
| An opposite example (looks difficult, solved easily):
> motorist drives from NY to troy at an average speed of 30 MPH,
> and from troy to NY at 60 MPH (back the same road).
< Assuming the trip at each direction begins at the same time of
< day, prove (1 line proof) that there exists a point on the road
< which the motorist visited twice at exactly the same time of day.
/Nitsan
|
| 1657.2 | attempt at answering .1 | STAR::ABBASI | Have you spelled checked today? | Sun Aug 30 1992 16:53 | 9 |
| imagine a motorist A leaves NY to TROY, and at same time a motorist B
leaves TROY to NY on the same road, then they must pass each others at
one point on the road, when this happens it will be offcourse same time
of the day for each of them. this answers .1
/Nasser
could not squeeze all the above on one line ;-)
|
| 1657.3 | Twist | HOBBLE::GERTLER | | Mon Aug 31 1992 11:49 | 7 |
| Here's an interesting twist:
If you travel from Troy to NY at 30 mph, how fast would you have
to drive the return leg such that the average speed for the round
trip is 60 mph?
-Slice
|
| 1657.4 | must be very fassssst ?! | STAR::ABBASI | Have you spelled checked today? | Mon Aug 31 1992 13:03 | 18 |
| I get that the return trip must take 0 units of time for the average
speed to be 60 MPH, this means the speed for the return part of the
trip must be infinity !
60 = total distance/total_time
= 2X/(t_1+t_2)
but 30= X/t_1
t_1= X/30
so 60= 2X/(X/30+t_2)
30= X/(X/30+t_2)
30= 30X/(X+30t_2)
1= X/(X+30t_2)
1= 1/(1+ 30 t_2/x)
so 30 t_2/X = 0
so t_2 = 0
/nasser
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