Question

current attempt in progress
an airplane makes a gradual $90^\circ$ turn while flying at a constant speed of 200 m/s. the process takes 20.0 seconds to complete. for this turn the magnitude of the average acceleration of the plane is:
$10 \\, \text{m/s}^2$
$20 \\, \text{m/s}^2$
$40 \\, \text{m/s}^2$
$0 \\, \text{m/s}^2$
$14 \\, \text{m/s}^2$

Explanation:

Step1: Define velocity vectors

Let initial velocity $\vec{v}_i = 200\hat{i} \, \text{m/s}$. After 90° turn, final velocity $\vec{v}_f = 200\hat{j} \, \text{m/s}$.

Step2: Calculate velocity change

Find $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = 200\hat{j} - 200\hat{i} \, \text{m/s}$.
Magnitude: $|\Delta\vec{v}| = \sqrt{(200)^2 + (-200)^2} = 200\sqrt{2} \, \text{m/s}$

Step3: Compute average acceleration

Use $\bar{a} = \frac{|\Delta\vec{v}|}{\Delta t}$, $\Delta t=20.0\,\text{s}$.
$\bar{a} = \frac{200\sqrt{2}}{20.0} = 10\sqrt{2} \approx 14 \, \text{m/s}^2$

Answer:

14 m/s²