Answer is: option1

-30

Solution:

The velocity function is given as:

\( v(t) = -t^3 + 6t^2 - 15t + 10 \)

Acceleration is the derivative of velocity:

\( a(t) = \frac{dv}{dt} = -3t^2 + 12t - 15 \)

To find critical points, set the derivative of \( a(t) \) to zero:

\( \frac{d}{dt}(-3t^2 + 12t - 15) = -6t + 12 = 0 \)

Solving for \( t \):

\( -6t + 12 = 0 \)

\( t = 2 \)

Since the interval is \( 0 \leq t \leq 5 \), we evaluate \( a(t) \) at \( t = 0, 2, 5 \).

• At \( t = 0 \):
  \( a(0) = -3(0)^2 + 12(0) - 15 = -15 \)
• At \( t = 2 \):
  \( a(2) = -3(2)^2 + 12(2) - 15 \)
  \( = -3(4) + 24 - 15 \)
  \( = -12 + 24 - 15 = -3 \)
• At \( t = 5 \):
  \( a(5) = -3(5)^2 + 12(5) - 15 \)
  \( = -3(25) + 60 - 15 \)
  \( = -75 + 60 - 15 = -30 \)

From the computed values:

\( a(0) = -15, \quad a(2) = -3, \quad a(5) = -30 \)

The minimum acceleration is -30, which occurs at \( t = 5 \).