To solve this problem, we'll use the following kinematic equation, which relates distance traveled $\Delta x$ to acceleration $a$, time $t$, and initial velocity $v_0$:

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

Given that:

• The go-kart starts from rest, so $v_0 = 0$ m/s.
• The acceleration $a = 1.77 \, \text{m/s}^2$.
• The time $t = 2.33 \, \text{s}$.

Substituting these values into the equation:

$\Delta x = 0 \cdot 2.33 + \frac{1}{2} \cdot 1.77 \cdot (2.33)^2$

Now let's calculate the value:

$\Delta x = \frac{1}{2} \cdot 1.77 \cdot 2.33^2$

$\Delta x = 0.885 \cdot 5.4289$

$\Delta x \approx 4.8054 \, \text{m}$

So, the go-kart travels approximately 4.81 meters during the 2.33 seconds.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How would the distance change if the acceleration were doubled?
2. What if the go-kart had an initial velocity? How would that affect the distance?
3. How would you calculate the final velocity of the go-kart after 2.33 seconds?
4. What would be the distance if the time was doubled while keeping the same acceleration?
5. How can you use the same formula to find the time if the distance and acceleration are known?

Tip: Remember that kinematic equations assume constant acceleration. If acceleration changes, more complex calculations are required.