To find the acceleration of the cheetah, we can use the following kinematic equation:

$a = \frac{{v_f - v_i}}{{t}}$

Where:

• $a$ is the acceleration,
• $v_f$ is the final velocity,
• $v_i$ is the initial velocity,
• $t$ is the time taken.

Given:

• $v_i = 0 \, \text{m/s}$ (since the cheetah starts from rest),
• $v_f = 30.0 \, \text{m/s}$,
• $t = 7.00 \, \text{s}$.

Now, substitute the values into the equation:

$a = \frac{{30.0 \, \text{m/s} - 0 \, \text{m/s}}}{{7.00 \, \text{s}}}$

$a = \frac{{30.0}}{{7.00}} = 4.29 \, \text{m/s}^2$

So, the acceleration of the cheetah is 4.29 m/s².

Would you like more details or have any questions about this problem?

Here are some related questions for you:

1. What is the final velocity of an object with a constant acceleration if the initial velocity is 0 m/s?
2. How does time affect the acceleration when the initial and final velocities are fixed?
3. What is the relationship between acceleration and the force applied to an object?
4. How would the cheetah's acceleration change if it took 10 seconds to reach the same speed?
5. How can you calculate the distance traveled by an object with constant acceleration?

Tip: When dealing with problems involving acceleration, always make sure to distinguish between initial and final velocities and the time taken to reach those velocities.