equations in 3 variable

Question

if 2 trains are moving in opposite direction then they take 10 seconds to cross each other . If they move in the same directions then they take  30 seconds to cross each other . if the length of the longer train is decreased by 50% then it takes 8 seconds less  to cross each other while moving in same direction. what is the time taken by longer train to cross atunnel 2 times its length , if the length of  longer train and shorter train differ by 25 meters .

Answer 1

For two trains A and B to pass one another they each have to travel their combined lengths to pass clear. When they are travelling in opposite directions, the time to cover this distance is given by:

₁t_o=(L_A+L_B)/(v_A+v_B)=10 secs, where L represent length and speed of the trains (subscript identifies which train). Speed is measured in metres/second.

When travelling in the same direction time to pass is given by:

₁t_s=(L_A+L_B)/(v_A-v_B)=30 secs.

If L_A>L_B then the changed length of L_A becomes L_A/2. It takes 30-8=22 seconds:

₂t_s=(L_A/2+L_B)/(v_A-v_B)=22 secs.

L_A-L_B=25m.

The length of the tunnel is 2L_A and to pass completely through the tunnel train A has to travel 2L_A+L_A=3L_A metres. Therefore, train takes 3L_A/v_A seconds to completely pass through the tunnel.

We have to assume that the speed of each train is constant for the whole exercise.

All distances and lengths can be written in terms of L_A, so L_B=L_A-25 metres. So we can rewrite the equations:

₁t_o=(2L_A-25)/(v_A+v_B)=10 secs, ₁t_s=(2L_A-25)/(v_A-v_B)=30 secs, ₂t_s=(3L_A/2-25)/(v_A-v_B)=22 secs.

We need t=3L_A/v_A, the time it takes for train A to pass through the tunnel.

(2L_A-25)/(v_A+v_B)=10, so 2L_A-25=10v_A+10v_B; 2L_A/v_A-25/v_A=10+10v_B/v_A; ⅔t-25/v_A=10+10v_B/v_A;

(2L_A-25)/(v_A-v_B)=30, so 2L_A-25=30v_A-30v_B; ⅔t-25/v_A=30-30v_B/v_A;

(3L_A/2-25)/(v_A-v_B)=22, so 3L_A-50=44v_A-44v_B; t-50/v_A=44-44v_B/v_A.

10+10v_B/v_A=30-30v_B/v_A; 40v_B/v_A=20, v_B/v_A=½, so we know v_B in terms of v_A. So we can substitute for v_B/v_A.

⅔t-25/v_A=10+5=15 (also ⅔t-25/v_A=30-15=15); t-50/v_A=44-22=22. We now have two equations and two unknowns:

⅔t-25/v_A=15; t-50/v_A=22.

4t/3-50/v_A=30; 50/v_A=4t/3-30=t-22, ⅓t=8, t=24 seconds is the time for train A to completely pass through the tunnel.