三元一次方程 + 普通力学

Better solution:

Person 1,2,3 take the car first, while 4,5 walk. After dropping off person 2 and 3, person 1 drive the back to pick up 4 & 5, while person 2 & 3 continue walking.

T1 = the time before dropping off 2 $ 3
T2 = the time between dropping off 2/3 and picking up 4/5
T3 = the time after picking up 4/5

L11 = 4 x T1 = distance by person 4/5 during T1.
L12 = 10 x T1 = distance by the car during T1
L21 = 4 x T2 = distance by person 4/5 during T2
L22 = 10 X T2 = distance by the car during T2
L23 = 2 x T2 = distance by person 2/3 during T2
L32 = 10 x T3 = distance by the car during T3
L33 = 2 x T3 = distance by 2/3 in T3

L11 + L21 = L22 + L23 + L33
L12 = L11 + L21 + L22
L12 + L23 + L33 = 1000

a) 4 x T1 + 4 x T2 = 10 x T2 + 2 x T2 + 2 x T3
b) 10 x T1 = 4 x T1 + 4 x T2 + 10 x T2
c) 10 x T1 + 2 x T1 + 2 x T2 = 1000

a) 2 x T1 - 4 x T2 - T3 = 0
b) 3 x T1 - 7 x T2 = 0 ==> 3 x T1 = 7 x T2
c) 6 x T1 + T2 = 500

replace 3 x T1 with 7 T2 in c)
14 x T2 + T2 = 500 => 15 x T2 = 500 ==> 3 x T2 = 100
T2 = 100/3

using b)
T1 = (7 x 100/3)/3 = 700/9

replace T1 and T2 in a)
T3 = 2 x T1 - 4 x T2 = 1400/9 - 400/3

T = T1 + T2 + T3
= 700/9 + 100/3 + 1400/9 - 400/3 = 2100/9 - 300/3
= 700/3 - 300/3 = 400 / 3
= 133.33

所有跟帖： 

• 五人不能同时到达终点？mean not a perfec solution -xyz100- ♂ (0 bytes) () 10/23/2008 postreply 11:22:35