Answer:
The initial distance between the trains is 1450 m.
Explanation:
In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.
Train B’s acceleration is [tex]1m/s^2[/tex] and it accelerated for 50 seconds.
[tex]a=1 m/s^2[/tex]
t=50 s
initial speed u=72km/h
we have to convert this speed into m/s
[tex]u=72 \times 5/18=20 m/s[/tex]
Distance covered in accelerating phase [tex]S=ut+1/2 at^2[/tex]
[tex]=20 \times 50+1/2 \times 1 \times 50^2[/tex]
[tex]=1000+1250=2250 m[/tex]
If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.
Here length of train A+length of train [tex]B=400+400=800 m[/tex]
Hence the initial distance between the trains = [tex]2250-800=1450 m[/tex]
