To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.
For this purpose we have that the constructive interference in waves can be expressed under the function

Where
a = Width of the slit
d = Distance of slit to screen
m = Number of order which represent the number of repetition of the spectrum
Angle between incident rays and scatter planes
At the same time the distance on the screen from the central point, would be

Where y = Represents the distance on the screen from the central point
PART A ) From the previous equation if we arrange to find the angle we have that



PART B) Equation both equations we have


Re-arrange to find a,


Answer:
a) 17.49 seconds
b) 13.12 seconds
c) 2.99 m/s²
Explanation:
a) Acceleration = a = 1.35 m/s²
Final velocity = v = 85 km/h = 
Initial velocity = u = 0
Equation of motion

Time taken to accelerate to top speed is 17.49 seconds.
b) Acceleration = a = -1.8 m/s²
Initial velocity = u = 23.61\ m/s
Final velocity = v = 0

Time taken to stop the train from top speed is 13.12 seconds
c) Initial velocity = u = 23.61 m/s
Time taken = t = 7.9 s
Final velocity = v = 0

Emergency acceleration is 2.99 m/s² (magnitude)
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration