Answer is: option2

Solution:

v(t) = e^t / t

Using the quotient rule:

v'(t) = (e^t·t − e^t·1) / t² = (e^t(t − 1)) / t²

Set the derivative equal to zero to find critical points:

v'(t) = 0    ⟹    (e^t(t − 1)) / t² = 0

Since e^t > 0 and t² > 0 for t > 0, the equation simplifies to:
t − 1 = 0    ⟹    t = 1

For t < 1 (e.g., t = 0.5):

v'(0.5) = (e^0.5(0.5 − 1)) / (0.5)² = Negative

The function is decreasing.

For t > 1 (e.g., t = 2):

v'(2) = (e²(2 − 1)) / 2² = Positive

The function is increasing.

Since the function changes from decreasing to increasing at t = 1, this point corresponds to a minimum.

The velocity v attains its minimum at t = 1.

Answer: B