Answer:
The shorter train has a length of 600m
Explanation:
The equation of motion of the first train can be written as:
[tex]x'=v't[/tex] (1)
v' = 108km/h = 30m/s
Furthermore, you can write the equation of motion of the second train as:
[tex]x=(200+l)+vt[/tex] = xo + vt (2)
v = 36km/h = 10m/s
where you have taken the initial position as measured from the front of the first train to the front of the second one. l is the length of the second train and 200 the separation of it respect to the first train.
For t = 40 both trains have the same position, that is, x=x'. Then, you equal equations (1) and (2), replace t=40, and solve the equation for l:
[tex]x'=x\\\\v't=200+l+vt\\\\(30m/s)(40s)=200+l+(10m/s)(40s)\\\\l=600m[/tex]
Hence, the first train has a length of 600 m
The train behind, or the second train, has twice the length of the first train:
[tex]l'=2l=2(600m)=1200m[/tex]
The shorter train has a length of 600m