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Sagot :
To find the component form of the velocity vector of the hand glider diving at [tex]$25$[/tex] mph in a direction [tex]$60^\circ$[/tex] below the horizontal, we can break this down into step-by-step components.
1. Understand the Problem:
- The speed of the hand glider is 25 mph.
- The angle of descent is [tex]$60^\circ$[/tex] below the horizontal.
2. Using Trigonometry:
- We need to find the horizontal ([tex]\(v_x\)[/tex]) and vertical ([tex]\(v_y\)[/tex]) components of the velocity.
- For an angle [tex]$\theta$[/tex] below the horizontal:
- The horizontal component [tex]\(v_x\)[/tex] of the velocity can be found using [tex]\(v_x = v \cdot \cos(\theta)\)[/tex].
- The vertical component [tex]\(v_y\)[/tex] of the velocity direction is downward (hence it will be negative), and can be found using [tex]\(v_y = v \cdot \sin(\theta)\)[/tex].
3. Calculate the Components:
- Given [tex]$\theta = -60^\circ$[/tex] (since it's below the horizontal).
- The speed [tex]\(v = 25 \, \text{mph}\)[/tex].
- First, we convert the angle into radians because standard trigonometric functions usually use radians. However, since we are moving forward with knowledge of calculating using degrees in practical terms, we keep it in degrees for simplicity.
[tex]\[ v_x = 25 \cdot \cos(-60^\circ) \][/tex]
[tex]\[ v_y = 25 \cdot \sin(-60^\circ) \][/tex]
4. Simplify the Computations:
- The cosine function is even, so [tex]\(\cos(-60^\circ) = \cos(60^\circ)\)[/tex].
- Similarly, the sine function is odd, so [tex]\(\sin(-60^\circ) = -\sin(60^\circ)\)[/tex].
Therefore:
[tex]\[ v_x = 25 \cdot \cos(60^\circ) \][/tex]
[tex]\[ v_y = 25 \cdot (-\sin(60^\circ)) \][/tex]
Using known values:
[tex]\[ \cos(60^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \][/tex]
5. Substitute and Solve:
[tex]\[ v_x = 25 \cdot \frac{1}{2} = 12.5 \][/tex]
[tex]\[ v_y = 25 \cdot -\frac{\sqrt{3}}{2} = -12.5 \sqrt{3} \][/tex]
Calculating [tex]\(v_y\)[/tex]:
[tex]\[ v_y \approx -21.65 \][/tex]
Thus, the component form of the velocity vector is approximately [tex]\(\left(12.5, -21.65\right)\)[/tex]. This corresponds exactly to the numerical solution provided earlier.
Given the answer choices, the correct match is:
[tex]\[ \left( \frac{25}{2}, -\frac{25 \sqrt{3}}{2} \right) \][/tex]
So the correct answer is:
[tex]\[ \left( \frac{25}{2}, -\frac{25 \sqrt{3}}{2} \right) \][/tex]
1. Understand the Problem:
- The speed of the hand glider is 25 mph.
- The angle of descent is [tex]$60^\circ$[/tex] below the horizontal.
2. Using Trigonometry:
- We need to find the horizontal ([tex]\(v_x\)[/tex]) and vertical ([tex]\(v_y\)[/tex]) components of the velocity.
- For an angle [tex]$\theta$[/tex] below the horizontal:
- The horizontal component [tex]\(v_x\)[/tex] of the velocity can be found using [tex]\(v_x = v \cdot \cos(\theta)\)[/tex].
- The vertical component [tex]\(v_y\)[/tex] of the velocity direction is downward (hence it will be negative), and can be found using [tex]\(v_y = v \cdot \sin(\theta)\)[/tex].
3. Calculate the Components:
- Given [tex]$\theta = -60^\circ$[/tex] (since it's below the horizontal).
- The speed [tex]\(v = 25 \, \text{mph}\)[/tex].
- First, we convert the angle into radians because standard trigonometric functions usually use radians. However, since we are moving forward with knowledge of calculating using degrees in practical terms, we keep it in degrees for simplicity.
[tex]\[ v_x = 25 \cdot \cos(-60^\circ) \][/tex]
[tex]\[ v_y = 25 \cdot \sin(-60^\circ) \][/tex]
4. Simplify the Computations:
- The cosine function is even, so [tex]\(\cos(-60^\circ) = \cos(60^\circ)\)[/tex].
- Similarly, the sine function is odd, so [tex]\(\sin(-60^\circ) = -\sin(60^\circ)\)[/tex].
Therefore:
[tex]\[ v_x = 25 \cdot \cos(60^\circ) \][/tex]
[tex]\[ v_y = 25 \cdot (-\sin(60^\circ)) \][/tex]
Using known values:
[tex]\[ \cos(60^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \][/tex]
5. Substitute and Solve:
[tex]\[ v_x = 25 \cdot \frac{1}{2} = 12.5 \][/tex]
[tex]\[ v_y = 25 \cdot -\frac{\sqrt{3}}{2} = -12.5 \sqrt{3} \][/tex]
Calculating [tex]\(v_y\)[/tex]:
[tex]\[ v_y \approx -21.65 \][/tex]
Thus, the component form of the velocity vector is approximately [tex]\(\left(12.5, -21.65\right)\)[/tex]. This corresponds exactly to the numerical solution provided earlier.
Given the answer choices, the correct match is:
[tex]\[ \left( \frac{25}{2}, -\frac{25 \sqrt{3}}{2} \right) \][/tex]
So the correct answer is:
[tex]\[ \left( \frac{25}{2}, -\frac{25 \sqrt{3}}{2} \right) \][/tex]
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