QuestionMay 24, 2025

A plane is flying at 300m/s in the y -direction. The wind is blowing at 20.0m/s in the x -direction. What is the magnitude of the velocity of the plane? overrightarrow (v)=[?]m/s overrightarrow (v)=sqrt ((v_(y))^2+(v_(x))^2)

A plane is flying at 300m/s in the y -direction. The wind is blowing at 20.0m/s in the x -direction. What is the magnitude of the velocity of the plane? overrightarrow (v)=[?]m/s overrightarrow (v)=sqrt ((v_(y))^2+(v_(x))^2)
A plane is flying at 300m/s
in the y -direction.
The wind is blowing at 20.0m/s
in the x -direction.
What is the magnitude of
the velocity of the plane?
overrightarrow (v)=[?]m/s
overrightarrow (v)=sqrt ((v_(y))^2+(v_(x))^2)

Solution
4.2(246 votes)

Answer

\overrightarrow{v} \approx 300.67 \, \text{m/s} Explanation 1. Identify Components The plane's velocity in the y-direction is v_y = 300 \, \text{m/s} and the wind's velocity in the x-direction is v_x = 20 \, \text{m/s}. 2. Apply Pythagorean Theorem Use the formula for magnitude of velocity: **\overrightarrow{v} = \sqrt{(v_y)^2 + (v_x)^2}**. 3. Calculate Magnitude Substitute the values: \overrightarrow{v} = \sqrt{(300)^2 + (20)^2} = \sqrt{90000 + 400} = \sqrt{90400}. 4. Compute Final Result Calculate the square root: \overrightarrow{v} \approx 300.67 \, \text{m/s}.

Explanation

1. Identify Components<br /> The plane's velocity in the y-direction is $v_y = 300 \, \text{m/s}$ and the wind's velocity in the x-direction is $v_x = 20 \, \text{m/s}$.<br /><br />2. Apply Pythagorean Theorem<br /> Use the formula for magnitude of velocity: **$\overrightarrow{v} = \sqrt{(v_y)^2 + (v_x)^2}$**.<br /><br />3. Calculate Magnitude<br /> Substitute the values: $\overrightarrow{v} = \sqrt{(300)^2 + (20)^2} = \sqrt{90000 + 400} = \sqrt{90400}$.<br /><br />4. Compute Final Result<br /> Calculate the square root: $\overrightarrow{v} \approx 300.67 \, \text{m/s}$.
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