Answer :
Answer:
The boat's velocity relative to Earth is 3.91 m/s.
Explanation:
Given;
speed of ferry due north, [tex]V_y[/tex] = 2.5 m/s
speed of the river to the east, [tex]V_x[/tex] = 3.0 m/s
The boat's velocity relative to Earth is given by;
[tex]V = \sqrt{V_x^2 + V_y^2} \\\\V = \sqrt{(3)^2+(2.5)^2}\\\\V = 3.91 \ m/s[/tex]
Therefore, the boat's velocity relative to Earth is 3.91 m/s.
The boat’s velocity relative to Earth be 3.91 m/s.
Given that,
- A ferry is crossing a river. If the ferry is headed due north with a speed of 2.5 m/s relative to the water.
- And the river’s velocity is 3.0 m/s to the east.
Based on the above information, the calculation is as follows:
[tex]= \sqrt{2.5^2 + 3.0^2} \\\\= \sqrt{6.25 + 9}\\\\= \sqrt{15.25}[/tex]
= 3.91 m/s
Therefore we can conclude that the boat’s velocity relative to Earth be 3.91 m/s.
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