To find the velocity of the torpedo in relation to a stationary observer on the ground, we need to add the velocity of the torpedo fired by the ship to the velocity of the ship's movement.
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Torpedo Velocity: The torpedo is fired due west at a speed of 250 km/h. So, in vector form, the velocity of the torpedo can be expressed as: \[ \vec{V}_{\text{torpedo}} = (-250, 0) \text{ km/h} \]
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Ship Velocity: The ship sails at 80 km/h in a direction of 30° south of west. We need to break this velocity into its x (west-east) and y (north-south) components.
- The angle 30° south of west means we measure 30° from the westward direction (negative x-axis) towards the south (negative y-axis): \[ V_{\text{ship}} = 80 \text{ km/h} \]
- The x-component (west) is: \[ V_{x,\text{ship}} = -80 \cos(30°) = -80 \times \frac{\sqrt{3}}{2} \approx -69.28 \text{ km/h} \]
- The y-component (south) is: \[ V_{y,\text{ship}} = -80 \sin(30°) = -80 \times \frac{1}{2} = -40 \text{ km/h} \]
Thus, the velocity of the ship can be expressed in vector form: \[ \vec{V}_{\text{ship}} = (-69.28, -40) \text{ km/h} \]
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Combined Velocity of the Torpedo: To find the actual velocity of the torpedo, we add the velocity vectors of the torpedo and the ship: \[ \vec{V}{\text{torpedo, combined}} = \vec{V}{\text{torpedo}} + \vec{V}{\text{ship}} \] \[ \vec{V}{\text{torpedo, combined}} = (-250, 0) + (-69.28, -40) = (-319.28, -40) \text{ km/h} \]
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Magnitude and Direction: Now, we need to calculate the magnitude of the combined velocity: \[ |\vec{V}_{\text{torpedo, combined}}| = \sqrt{(-319.28)^2 + (-40)^2} \] \[ = \sqrt{101922.59 + 1600} \] \[ = \sqrt{103522.59} \approx 321.38 \text{ km/h} \]
To find the direction (angle south of west), we use the tangent function: \[ \tan(\theta) = \frac{|y|}{|x|} = \frac{40}{319.28} \] \[ \theta = \tan^{-1}\left(\frac{40}{319.28}\right) \approx \tan^{-1}(0.1259) \approx 7.13° \]
Thus, the velocity of the torpedo with respect to the ground is approximately 321.38 km/h at 7.13° south of west.
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