Answer is: option3

2.22

Solution:

The average velocity over the interval \([1,4]\) is given by:

\[ V_{\text{avg}} = \frac{s(4) - s(1)}{4 - 1} \]

where \( s(t) = \frac{9t^2}{t^2 + 2} \).

Compute \( s(4) \): \[ s(4) = \frac{9(4)^2}{(4)^2 + 2} = \frac{9(16)}{16 + 2} = \frac{144}{18} = 8 \] Compute \( s(1) \): \[ s(1) = \frac{9(1)^2}{(1)^2 + 2} = \frac{9(1)}{1 + 2} = \frac{9}{3} = 3 \] Compute \( V_{\text{avg}} \): \[ V_{\text{avg}} = \frac{8 - 3}{4 - 1} = \frac{5}{3} \approx 1.67 \] The instantaneous velocity is given by the derivative \( s'(t) \): \[ s(t) = \frac{9t^2}{t^2 + 2} \] Using the quotient rule: \[ s'(t) = \frac{(18t)(t^2 + 2) - (9t^2)(2t)}{(t^2 + 2)^2} \] \[ = \frac{18t(t^2 + 2) - 18t^3}{(t^2 + 2)^2} \] \[ = \frac{18t^3 + 36t - 18t^3}{(t^2 + 2)^2} \] \[ = \frac{36t}{(t^2 + 2)^2} \] Solve for \( t \) when \( s'(t) = V_{\text{avg}} \) \[ \frac{36t}{(t^2 + 2)^2} = \frac{5}{3} \]

Using Desmos, we get:

\[ t = 2.21873 \] Thus, the correct answer is: \[ \boxed{2.22} \]