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 :
[tex]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[/tex]
So, the minimum distance is 2083.33 m
