Question

Two trains run in the same direction. The one behind has double the length of the one ahead. The initial distance of separation = 200 m. The speeds are(a) train in front = 36 km/h(b) train behind = 108 km/h. If it took 40 seconds for the longer train to overtake the shorter one , find the length of the shorter train.

Answer 1

Answer:

The shorter train has a length of 600m

Explanation:

The equation of motion of the first train can be written as:

$x'=v't$  (1)

v' = 108km/h = 30m/s

Furthermore, you can write the equation of motion of the second train as:

$x=(200+l)+vt$  = 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:

$x'=x\\\\v't=200+l+vt\\\\(30m/s)(40s)=200+l+(10m/s)(40s)\\\\l=600m$

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:

$l'=2l=2(600m)=1200m$

The shorter train has a length of 600m