Hi there!
We can begin by calculating the distance remaining after the reaction time.
Δd = vt
Calculate the distance traveled within this time:
Δd = (16)(.79) = 12.64 m
Subtract from the total distance:
150 - 12.64 = 137.66 m remaining
We can use the following equation to solve for the acceleration necessary:
vf² = vi² + 2ad, where vf = 0 since the train will have slowed down to rest.
Rearrange in terms of "a":
0 = vi² + 2ad
(-vi²) = 2ad
(-vi²)/2d = a
Plug in the given values:
(-(16²))/2(137.66) = a
-256/275.32 = -.9298 m/s²