Answer is: option4

\( 1 < t < 4 \) and \( t > 7 \)

Solution:

The position function is given by:

s(t) = t³ - 12t² + 21t + 10

Velocity is the first derivative of position:

v(t) = ds/dt = 3t² - 24t + 21

Acceleration is the derivative of velocity:

a(t) = dv/dt = 6t - 24

The speed of the particle is increasing when:

v(t) and a(t) have the same sign (both positive or both negative).

Find Critical Points of Velocity

Set v(t) = 0:

3t² - 24t + 21 = 0

Thus, velocity changes sign at t = 1 and t = 7.

Set a(t) = 0:

6t - 24 = 0

t = 4

Thus, acceleration changes sign at t = 4.

For t < 1:

• Pick t = 0, then v(0) = 3(0)² - 24(0) + 21 = 21 (positive)
• Pick t = 0, then a(0) = 6(0) - 24 = -24 (negative)
• v(t) and a(t) have opposite signs → speed is decreasing.

For 1 < t < 4:

• Pick t = 2, then v(2) = 3(2)² - 24(2) + 21 = -15 (negative)
• Pick t = 2, then a(2) = 6(2) - 24 = -12 (negative)
• v(t) and a(t) have the same sign → speed is increasing.

For 4 < t < 7:

• Pick t = 5, then v(5) = 3(5)² - 24(5) + 21 = -24 (negative)
• Pick t = 5, then a(5) = 6(5) - 24 = 6 (positive)
• v(t) and a(t) have opposite signs → speed is decreasing.

For t ≥ 7:

• Pick t = 8, then v(8) = 3(8)² - 24(8) + 21 = 21 (positive)
• Pick t = 8, then a(8) = 6(8) - 24 = 24 (positive)
• v(t) and a(t) have the same sign → speed is increasing.

Conclusion:

From our analysis, speed is increasing in the intervals:

1 < t < 4 and t > 7