To find the cheetah's acceleration, we can use the formula for acceleration, which is given by:

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

Where:

• $a$ is the acceleration,
• $\Delta v$ is the change in velocity,
• $\Delta t$ is the time interval.

Given:

• Initial velocity ($v_0$) = 0 m/s (since the cheetah starts from rest),
• Final velocity ($v_f$) = 30.0 m/s,
• Time ($\Delta t$) = 7.19 s.

Now, plug the values into the formula:

$a = \frac{{30.0 \, \text{m/s} - 0 \, \text{m/s}}}{7.19 \, \text{s}}$

$a = \frac{{30.0}}{7.19} \, \text{m/s}^2$

Let me calculate this for you.

The cheetah's acceleration is approximately $4.17 \, \text{m/s}^2$.

Would you like more details or have any other questions?

Here are five related questions to expand on this topic:

1. What would be the displacement of the cheetah during this time period?
2. How much time would it take for the cheetah to reach 50.0 m/s at the same acceleration?
3. What is the final speed of the cheetah if it accelerates for 10 seconds?
4. What force does the cheetah exert if its mass is 60 kg and it accelerates at this rate?
5. How does the cheetah's acceleration compare to that of a car?

Tip: Always ensure units are consistent when applying formulas in physics to avoid calculation errors.