Question

A train can speed up at a uniform rate of 0.15 m/s2. In what minimum distance can

Answer 1

Answer:

d = 2083.33 m

Explanation:

Given that,

Acceleration of the train, a = 0.15 m/s²

The initial speed of the car, u = 0\

Final velocity, v = 25 m/s

We need to find the minimum distance covered by the train. Let it is d. Using third equation of kinematics as follows :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(25)^2-0}{2\times 0.15}\\\\d=2083.33\ m$

So, the minimum distance is 2083.33 m