# Two trains A and B 100 km part are travelling towards each other with starting speed of 50 km/hr for both. The train A accelerating at 18 ?

Two trains A and B 100 km part are travelling towards each other with starting speed of 50 km/hr for both. The train A accelerating at 18 km/hr² and B decelerating at 18 km/hr². At what distance from the starting point of A,  will they meet?

Relevance

Let the two trains meet each other after t hour, the distance covered by both in t hour shall be 100 km,

For train A:

=>by s = ut + 1/2at^2

=>s1 = 50t + 1/2 x 18 x t^2

=>s1 = 50t + 9t^2 ---------------------(i)

For train B:

=>by s = ut - 1/2at^2

=>s1 = 50t - 1/2 x 18 x t^2

=>s1 = 50t - 9t^2 ---------------------(ii)

But s1 + s2 = 100

=>50t + 9t^2 + 50t - 9t^2 = 100

=>100t = 100

=>t = 1 hour

By (i): s1 = 50 x 1 + 9 x (1)^2

=>s1 = 59 km

• The relative speed between them at the beginning was 100 km / hr. The acceleration of both are the same so the relative speed remains 100 km/hr at all times.  Therefore time = distance / speed = 100/ 100 = 1 hour.  and the distance = 50 + 1/2 * 18 * t =  59 km

• Imagine yourself flying from A to B, from the midpoint, with an acceleration of 18 km/hr^2. Then you will see A catching op with you at a constant speed of 50km/h and B approaching you with a constant speed of 50km/h. (by accelerating yourself you will have nullified their acceleration/deceleration relative to you. This makes it an easier problem!) All in all both will meet you after 1 hour therefore. In that hour you have moved a distance

s = 1/2 a t^2 = 9 km.

So A will have covered 59 km (Your midway starting point + your 9 km)