A train travel on a straight track passing signal A at 20ms-¹.it accelerate uniformly at 2ms-² and reaches signal B 100m farther than A.AT B the velocity of train in ms-¹ is?

To solve this problem, we can use the equations of motion for uniformly accelerated motion.

Let's denote the initial velocity of the train as u, the acceleration as a, the distance between signals A and B as d, and the final velocity at signal B as v.

Given:
Initial velocity, u = 20 m/s
Acceleration, a = 2 m/s^2
Distance between signals A and B, d = 100 m

We know that the final velocity v can be calculated using the equation:
v^2 = u^2 + 2ad

Substitute the given values into the equation:
v^2 = (20)^2 + 2(2)(100)
v^2 = 400 + 400
v^2 = 800
v = √800
v ≈ 28.28 m/s

Therefore, the velocity of the train at signal B is approximately 28.28 m/s.

Initial velocity, u = 20 m/s
Acceleration, a = 2 m/s²
Distance travelled, s = 100 m

Using the first equation of motion:
v² - u² = 2as
v² - 20² = 2 - 2 - 100
v² = 20² + 400
v² = 800
v = √800
v = 28.28 m/s

Therefore, the velocity of the train at B is 28.28 m/s.