Answer is: option2

\( 3 \)

Solution:

Velocity Function:

\[ v(t) = -t^3 e^{-t} \]

Differentiating:

\[ v'(t) = (-3t^2 e^{-t}) + (-t^3 (-e^{-t})) \]

\[ = -3t^2 e^{-t} + t^3 e^{-t} \]

\[ = e^{-t} (-3t^2 + t^3) \]

Finding Critical Points:

Setting \( v'(t) = 0 \):

\[ e^{-t} (t^3 - 3t^2) = 0 \]

Since \( e^{-t} \neq 0 \), solving:

\[ t^3 - 3t^2 = 0 \]

\[ t^2 (t - 3) = 0 \]

\[ t^2 = 0 \quad \text{or} \quad t - 3 = 0 \]

\[ t = 0 \quad \text{or} \quad t = 3 \]

Confirming Minimum with Second Derivative:

\[ v''(t) = \frac{d}{dt} [e^{-t} (t^3 - 3t^2)] \]

Using the product rule:

\[ v''(t) = e^{-t} (3t^2 - 6t) + (-e^{-t}) (t^3 - 3t^2) \]

\[ = e^{-t} [(3t^2 - 6t) - (t^3 - 3t^2)] \]

\[ = e^{-t} (-t^3 + 6t^2 - 6t) \]

Evaluating at \( t = 3 \):

\[ v''(3) = e^{-3} (-27 + 54 - 18) \]

\[ = e^{-3} (9) \]

Since \( v''(3) > 0 \), \( t = 3 \) is a local minimum.