A cheetah can accelerate from rest to a speed of $30.0m/s$ in $7.00s$. What is its acceleration?

• A. $\frac{7.00}{30.0}m/s^2$ (approximately $0.233m/s^2$)
• B. $30.0m/s^2$
• C. $7.00m/s^2$
• D. $\frac{30.0}{7.00}m/s^2$ (approximately $4.29m/s^2$)

Answer: (D)

Verified step by step guidance

1. Identify the known quantities: the initial velocity \(v_0 = 0\ \text{m/s}\) (since the cheetah starts from rest), the final velocity \(v = 30.0\ \text{m/s}\), and the time interval \(t = 7.00\ \text{s}\) over which the acceleration occurs.
2. Recall the formula for constant acceleration when starting from rest: \(a = \frac{v - v_0}{t}\). Since \(v_0 = 0\), this simplifies to \(a = \frac{v}{t}\).
3. Substitute the known values into the formula: \(a = \frac{30.0\ \text{m/s}}{7.00\ \text{s}}\).
4. Perform the division to find the acceleration value (do not calculate the final number here, just set up the expression).
5. Interpret the result: the acceleration \(a\) will be in units of meters per second squared (\(\text{m/s}^2\)), representing how quickly the cheetah increases its speed each second.