To determine the position of the object at $t = 3$ seconds, we can use the kinematic equation for motion under constant acceleration:

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$

Where:

• $x(t)$ is the position at time $t$,
• $x_0$ is the initial position (assumed to be 0 if not provided),
• $v_0$ is the initial velocity (3 m/s),
• $a$ is the acceleration (12 m/s²),
• $t$ is the time (3 seconds).

Given:

• $x_0 = 0$ (assuming the initial position is zero),
• $v_0 = 3$ m/s,
• $a = 12$ m/s²,
• $t = 3$ seconds.

Let's calculate the position:

$x(3) = 0 + 3 \times 3 + \frac{1}{2} \times 12 \times (3)^2$ $x(3) = 9 + \frac{1}{2} \times 12 \times 9$ $x(3) = 9 + 54 = 63 \, \text{meters}$

The position of the object at 3 seconds is 63 meters.

Would you like any further explanation or details? Here are some related questions to explore:

1. What would the position be if the initial velocity were 0 m/s?
2. How would the position change if the acceleration was doubled?
3. What would the velocity of the object be at 3 seconds?
4. How long would it take for the object to reach a velocity of 30 m/s?
5. How would you calculate the displacement of the object over 5 seconds?

Tip: Kinematic equations are very useful for solving motion problems involving constant acceleration.