To find the displacement during the time interval, we can use one of the kinematic equations. The equation that relates displacement ($s$), initial velocity ($v_0$), final velocity ($v$), and acceleration ($a$) is:

$v^2 = v_0^2 + 2as$

We can rearrange this equation to solve for displacement ($s$):

$s = \frac{v^2 - v_0^2}{2a}$

Now, let's plug in the given values:

• $v = 10.4 \, \text{m/s}$
• $v_0 = 5.20 \, \text{m/s}$
• $a = 4.45 \, \text{m/s}^2$

$s = \frac{(10.4 \, \text{m/s})^2 - (5.20 \, \text{m/s})^2}{2 \times 4.45 \, \text{m/s}^2}$

$s = \frac{108.16 \, \text{m}^2/\text{s}^2 - 27.04 \, \text{m}^2/\text{s}^2}{8.90 \, \text{m/s}^2}$

$s = \frac{81.12 \, \text{m}^2/\text{s}^2}{8.90 \, \text{m/s}^2}$

$s = 9.12 \, \text{m}$

So, the displacement during the time interval is 9.12 meters.

Would you like more details or have any questions? Here are some related questions:

1. How would you calculate the time interval for this motion?
2. What would be the displacement if the acceleration were different?
3. How would you find the displacement if the final velocity were unknown?
4. How does the direction of acceleration affect displacement?
5. What if the initial velocity were negative—how would that change the calculation?

Tip: Always double-check the units when plugging values into kinematic equations to ensure consistency.