To find the acceleration, we can use the formula:

$a = \frac{{\Delta v}}{{\Delta t}}$

where:

• $\Delta v$ is the change in velocity.
• $\Delta t$ is the time taken for the change.

Given:

• Final velocity, $v = 31 \, \text{m/s}$
• Initial velocity, $u = 0 \, \text{m/s}$ (since it starts from rest)
• Time, $t = 7.6 \, \text{s}$

Now, calculate the acceleration:

$a = \frac{31 \, \text{m/s} - 0 \, \text{m/s}}{7.6 \, \text{s}} = \frac{31 \, \text{m/s}}{7.6 \, \text{s}} \approx 4.08 \, \text{m/s}^2$

So, the acceleration is 4.08 m/s².

Would you like further details or have any questions?

Here are some related questions:

1. How would the acceleration change if the time was doubled?
2. What would be the distance covered by the cheetah in these 7.6 seconds?
3. How does this acceleration compare to that of a car?
4. What is the average speed of the cheetah during the acceleration?
5. What forces act on the cheetah during this acceleration?

Tip: Always ensure your units are consistent when performing calculations.