Johnny left point A, walking North at a steady velocity V km/h . At the moment that he was at a distance of 6 km from A, a biker Tom left A to bring Johnny back, his speed being 9 km/h higher than Johnny's.
When they met they turned back and travelled South at 4 km/h.
For what value of V the duration of Johnny's trip would be minimal?
The distance traveled outbound by the second rider (Tom), and therefore also by Johnny, is given by
d = (v+9)(d/v - 6/v)
so
d(1 - (v+9)/v) = -(6/v)(v+9)
d = 6(v+9)/(v+9-v) = 2(v+9)/9
t1 = 2(v+9)/(9v); t2 = 2(v+9)/36
t' = (18v - 18(v+9)) / (81v^2) + 1/18
(18(v+9) - 18v) / (81v^2) = 1/18
18*9/(81v^2) = 1/18
81v^2 = 18^2 * 9
v^2 = 36
v = 6
so the duration of the trip is minimal when Johnny traveled at 6 km/hr.
The greatest distance achieved was 30/9 = 3.33... km. Going out it took Johnny 5/9 hr and returning it took 15/18 hr for a total of 1.3888... hr.
Verification of the calculus minimum-finding:
DEFDBL A-Z
CLS
FOR v = 4 TO 8 STEP .1#
t = 2 * (v + 9) / (9 * v) + 2 * (v + 9) / 36
PRINT USING "##.# ###.######"; v; t
NEXT v
v time
4.0 1.444444
4.1 1.437805
4.2 1.431746
4.3 1.426227
4.4 1.421212
4.5 1.416667
4.6 1.412560
4.7 1.408865
4.8 1.405556
4.9 1.402608
5.0 1.400000
5.1 1.397712
5.2 1.395726
5.3 1.394025
5.4 1.392593
5.5 1.391414
5.6 1.390476
5.7 1.389766
5.8 1.389272
5.9 1.388983
6.0 1.388889
6.1 1.388980
6.2 1.389247
6.3 1.389683
6.4 1.390278
6.5 1.391026
6.6 1.391919
6.7 1.392952
6.8 1.394118
6.9 1.395411
7.0 1.396825
7.1 1.398357
7.2 1.400000
7.3 1.401750
7.4 1.403604
7.5 1.405556
7.6 1.407602
7.7 1.409740
7.8 1.411966
7.9 1.414276
8.0 1.416667
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Posted by Charlie
on 2011-11-23 13:09:36 |
solution
