You are at the train station in downtown Riverside when a Me

Youe atthe train station, in downtown Riverside 4. You are at the train station, in downtown Riverside, when a Metrolink train arrives. You realize the train is slowing down, and you measure that the first car takes 1.29 seconds to pass you while the second car takes 1.76 seconds to pass you. The cars are each 85.0 feet long (including the gap between cars.) Assuming the train slows at a constant rate, find the: (a) magnitude of its acceleration; (b) total time, from when the front of the train was level with you, it takes the train to stop; (c) distance from you to the front of the train when it has stopped. (Hint: analyze the motion of the front of the first car.)

Solution

For the first car,
x = (vo)(t) + (1/2)at²
85.0 ft = (vo)(1.29 s) + (1/2)(a)(1.29 s)²
85.0 = 1.29(vo) + 0.83205a [1]

Velocity when second car starts to pass:
(v1) = (vo) + at
(v1) = (vo) + a(1.29 s)

For the second car,
x = (v1)(t) + (1/2)at²
85.0 ft = [(vo) + a(1.29 s)](2.04 s) + (1/2)(a)(2.04 s)²
85.0 = 2.04(vo) + 4.7124a [2]

Solve [1] for (vo), substitute into [2], and solve for a.
85.0 - 0.83205a = 1.29(vo)
(vo) = 65.89 - 0.645a

85.0 = 2.04(65.89 - 0.645a) + 4.7124a
85.0 = 134.4 - 1.316a + 4.7124a
-49.4 = 3.3964a
a = -14.5 ft/s² (magnitude: 14.5 ft/s²)

(b)
From part (a) find the initial velocity of the train.
(vo) = 65.89 - 0.645(-14.5 ft/s²)
(vo) = 75.24 ft/s

v = (vo) + at
0 = 75.24 ft/s + (-14.5 ft/s²)(t)
t = 5.19 s

(c)
2ax = v² - (vo)²
2(-14.5 ft/s²)(x) = 0 - (75.24 ft/s)²
x = 195 ft