A subway train starting from rest leaves a station with a constant acceleration. At the end of 5.24 s, it is moving at 13.4144 m/s.

What is the train’s displacement in the first 3.42696 s of motion?

v = at

13.4144 = 5.24a
a = 2.56

s = 1/2 at^2
s = .5 * 2.56 * 3.42696^2
s = 15.032390197

since you seem to like decimal places! :-)

To find the train's displacement in the first 3.42696 seconds of motion, we can use the equation for displacement with constant acceleration:

displacement = initial velocity * time + (1/2) * acceleration * time^2

We are given the following information:
Initial velocity (u) = 0 m/s (starting from rest)
Time (t) = 3.42696 s
Final velocity (v) = 13.4144 m/s

We need to find the acceleration (a). To find the acceleration, we can use the equation:

acceleration = (final velocity - initial velocity) / time

Substituting the given values:

acceleration = (13.4144 m/s - 0 m/s) / 5.24 s

acceleration ≈ 2.5586 m/s^2

Now we can find the displacement using the formula:

displacement = initial velocity * time + (1/2) * acceleration * time^2

Substituting the values:

displacement = 0 m/s * 3.42696 s + (1/2) * 2.5586 m/s^2 * (3.42696 s)^2

displacement ≈ 0 + 1/2 * 2.5586 m/s^2 * 11.75 s^2

displacement ≈ 1.2793 m/s^2 * 11.75 s^2

displacement ≈ 15.047 m

Therefore, the train's displacement in the first 3.42696 s of motion is approximately 15.047 meters.