1. **Problem statement:** We need to find the instantaneous velocity of an object moving in a straight line at time $t=3$ seconds, given its position function $$s(t) = 8\sqrt{t+1} - 8.$$\n\n2. **Formula and concept:** Instantaneous velocity is the derivative of the position function with respect to time: $$v(t) = \frac{ds}{dt}.$$\n\n3. **Rewrite the position function:** Note that $$\sqrt{t+1} = (t+1)^{\frac{1}{2}}.$$ So, $$s(t) = 8(t+1)^{\frac{1}{2}} - 8.$$\n\n4. **Differentiate $s(t)$:** Using the chain rule, $$\frac{d}{dt} (t+1)^{\frac{1}{2}} = \frac{1}{2}(t+1)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2\sqrt{t+1}}.$$\nTherefore, $$v(t) = \frac{ds}{dt} = 8 \cdot \frac{1}{2\sqrt{t+1}} - 0 = \frac{8}{2\sqrt{t+1}} = \frac{4}{\sqrt{t+1}}.$$\n\n5. **Evaluate at $t=3$ seconds:** $$v(3) = \frac{4}{\sqrt{3+1}} = \frac{4}{\sqrt{4}} = \frac{4}{2} = 2.$$\n\n6. **Interpretation:** The instantaneous velocity at 3 seconds is 2 meters per second.\n\n**Final answer:** The velocity at $t=3$ seconds is $2$ m/s, which corresponds to option A.