To find the distance traveled by the object, we can use the following kinematic equation for uniformly accelerated motion:

$d = v_0 t + \frac{1}{2} a t^2$

Where:

• $d$ is the distance traveled (what we're solving for),
• $v_0$ is the initial velocity (which is 0 since the object starts at rest),
• $a$ is the acceleration ($3 \, \text{m/s}^2$),
• $t$ is the time ($4 \, \text{seconds}$).

Since the initial velocity $v_0 = 0$, the formula simplifies to:

$d = \frac{1}{2} a t^2$

Now, plug in the given values:

$d = \frac{1}{2} \times 3 \times (4)^2$

$d = \frac{1}{2} \times 3 \times 16$

$d = \frac{1}{2} \times 48 = 24 \, \text{meters}$

So, the object has traveled 24 meters.

Would you like a deeper explanation on any of the concepts?

Here are 5 related questions to explore further:

1. What would happen if the acceleration changed over time?
2. How does the object's final velocity relate to the distance it traveled?
3. What if the object had an initial velocity instead of starting at rest?
4. How do you calculate distance if the acceleration is negative?
5. What is the significance of the factor $\frac{1}{2}$ in the kinematic equation?

Tip: In uniformly accelerated motion, the distance traveled depends on the square of the time. Doubling the time will quadruple the distance, assuming constant acceleration.