The question asks to find the velocity of a particle at $t = 3$ seconds given its distance $s = t^3 + 2t^2 + 3t + 4$.

1. Problem Description

The question asks to find the velocity of a particle at $t = 3$ seconds given its distance $s = t^3 + 2t^2 + 3t + 4$.

2. Solution Steps

To find the velocity, we need to differentiate the distance $s$ with respect to time $t$.

$s = t^3 + 2t^2 + 3t + 4$

Differentiating with respect to $t$:

$v = \frac{ds}{dt} = 3t^2 + 4t + 3$

Now, we substitute $t = 3$ into the expression for the velocity:

$v(3) = 3(3)^2 + 4(3) + 3$

$v(3) = 3(9) + 12 + 3$

$v(3) = 27 + 12 + 3$

$v(3) = 42$ m/s

3. Final Answer

(a) 42 m/s