42. In this solution we elect to wait until the last step to convert to SI units. Constant
acceleration is indicated, so use of Table 21 is permitted. We start with Eq. 217 and
denote the train’s initial velocity as
v
t
and the locomotive’s velocity as
v
A
(which is also
the final velocity of the train, if the rearend collision is barely avoided). We note that the
distance
Δ
x
consists of the original gap between them
D
as well as the forward distance
traveled during this time by the locomotive
v t
A
. Therefore,
v
v
x
t
D
v t
t
D
t
v
t
+
=
=
+
=
+
A
A
A
2
Δ
.
We now use Eq. 211 to eliminate time from the equation. Thus,
v
v
D
v
v
a
v
t
t
+
=
−
+
A
A
A
2
b
g
/
which leads to
a
v
v
v
v
v
D
D
v
v
t
t
t
=
+
−
F
H
G
I
K
J
−
F
H
G
I
K
J
= −
−
A
A
A
A
2
1
2
2
b
g
.
Hence,
a
= −
−
F
H
G
I
K
J
= −
1
2 0 676
29
161
12888
2
2
(
.
km)
km
h
km
h
km / h
which we convert as follows:
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 Spring '06
 Stoler
 Physics, Acceleration, km km

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