Let's solve the problem based on the given information.

Given:

• Initial velocity $u = 0$ (starts from rest)
• Acceleration $a = 3.00 \, \text{m/s}^2$
• Time $t = 30.0 \, \text{s}$

We will use the following kinematic equations:

1. To find the distance (part a): $s = ut + \frac{1}{2} a t^2$ Since $u = 0$, this simplifies to: $s = \frac{1}{2} a t^2$ Substituting the values: $s = \frac{1}{2} \times 3.00 \times (30.0)^2 = \frac{1}{2} \times 3.00 \times 900 = 1350 \, \text{m}$ So, the airplane moved 1350 meters.

2. To find the final velocity (part b): $v = u + at$ Again, $u = 0$, so: $v = a t = 3.00 \times 30.0 = 90.0 \, \text{m/s}$ So, the airplane's speed when it took off was 90.0 m/s.

Do you want further details on this solution or have any questions?

Here are 5 related questions for practice:

1. If the airplane accelerated at $4.00 \, \text{m/s}^2$, how far would it travel in 30 seconds?
2. How much time would it take the airplane to reach a speed of $120 \, \text{m/s}$ with the same acceleration?
3. What is the final velocity of the airplane if the acceleration lasts for 40 seconds?
4. What is the airplane's velocity after 20 seconds of acceleration?
5. If the airplane traveled 2000 meters before takeoff, what was its acceleration?

Tip: When solving kinematic problems, it's helpful to first list all known variables and choose the appropriate equation based on those variables.