To find the displacement of the airplane during its acceleration, we can use the kinematic equation for displacement:

$d = v_i t + \frac{1}{2} a t^2$

Where:

• $d$ is the displacement,
• $v_i$ is the initial velocity (which is 0 since the airplane starts from rest),
• $a$ is the acceleration ($2.90 \, \text{m/s}^2$),
• $t$ is the time ($25.92 \, \text{s}$).

Since the initial velocity $v_i = 0$, the equation simplifies to:

$d = \frac{1}{2} a t^2$

Now, plugging in the values:

$d = \frac{1}{2} (2.90 \, \text{m/s}^2) (25.92 \, \text{s})^2$

Let's calculate that.# Given values a = 2.90 # acceleration in m/s^2 t = 25.92 # time in seconds

Using the equation for displacement: d = 1/2 * a * t^2

displacement = 0.5 * a * t**2 displacementThe displacement of the airplane during the 25.92 seconds of acceleration is approximately $974.18 \, \text{meters}$.

Would you like further details or have any additional questions?

Here are 5 related questions to expand your understanding:

1. How would the displacement change if the acceleration was doubled?
2. What would be the final velocity of the airplane after this acceleration period?
3. How would the displacement change if the time was reduced to 15 seconds?
4. If the airplane continued accelerating for an additional 10 seconds, what would its displacement be then?
5. How can we calculate the distance if the airplane had an initial velocity?

Tip: When solving kinematic equations, always ensure the correct substitution of units to avoid calculation errors.