Read Example 3.36 in the textbook Links to an external site..

Suppose the position of a train travelling along a straight track is given by
s
(
t
)
=
t
3
−
6
t
2
+
12
t
where
t
≥
0
.

At what time is the train standing still?

(Answer with just a numerical answer e.g. 5, rather than "At t=5".)

Question

Read Example 3.36 in the textbook Links to an external site..

Suppose the position of a train travelling along a straight track is given by 
s
(
t
)
=
t
3
−
6
t
2
+
12
t
  where 
t
≥
0
.

At what time is the train standing still?

(Answer with just a numerical answer e.g. 5, rather than "At t=5".)

BlackTom AI · Accepted Answer

The problem gives the position function s(t) = t^3 − 6 t^2 + 12 t for t ≥ 0, and asks for the time when the train is standing still.
First, understand that standing still means zero velocity, which is the time derivative of position: v(t) = ds/dt.
Compute the derivative: ds/dt = d/dt (t^3 − 6 t^2 + 12 t) = 3 t^2 − 12 t + 12.
Factor the derivative to find where it equals zero: 3 t^2 − 12 t + 12 = 3 (t^2 − 4 t + 4) = 3 (t − 2)^2.
Set velocity to zero and solve for t: 3 (t − 2)^2 = 0 implies (t − 2)^2 = 0, so t = 2.
Check the domain: t ≥ 0, and t = 2 satisfies this condition, so it is valid.
Note that because the derivative is a squared term, there is a repeated root at t = 2, which means the train’s velocity is zero only at that instant (not for an interval).
Since the prompt asks for a numerical answer and not a formatted statement, the final value to report is 2.

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