Finding the resultant velocity?

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SUMMARY

The resultant velocity of a boat heading north at 4 miles per hour across a river with an eastward current of 3 miles per hour is calculated to be 5 miles per hour at an angle of approximately 53.13° to the bank. This is derived using the Pythagorean theorem, where the resultant velocity is the hypotenuse of a right triangle formed by the boat's velocity and the current. To find the angle, the cosine inverse function is applied, specifically cos-1(3/5). A diagram is recommended for visualizing the problem.

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Math9999

Homework Statement


A boat heads north across a river at a rate of 4 miles per hour. If the current is flowing east at a rate of 3 miles per hour, find the resultant velocity of the boat.

Homework Equations


None.

The Attempt at a Solution


I know that the resultant velocity is the hypotenuse, which means that 4^2+3^2=v^2, where v=5 mi/h. And that's only part of the answer. The whole answer in the textbook says 5 mi/h at an angle of ≈53.13° to the bank. But how do I find the angle to the bank?
 
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Math9999 said:

Homework Statement


A boat heads north across a river at a rate of 4 miles per hour. If the current is flowing east at a rate of 3 miles per hour, find the resultant velocity of the boat.

Homework Equations


None.

The Attempt at a Solution


I know that the resultant velocity is the hypotenuse, which means that 4^2+3^2=v^2, where v=5 mi/h. And that's only part of the answer. The whole answer in the textbook says 5 mi/h at an angle of ≈53.13° to the bank. But how do I find the angle to the bank?
Not without additional assumptions or conditions. You need to interpret "across the river" as perpendicular to the banks.
 
Math9999 said:

Homework Statement


A boat heads north across a river at a rate of 4 miles per hour. If the current is flowing east at a rate of 3 miles per hour, find the resultant velocity of the boat.

Homework Equations


None.

The Attempt at a Solution


I know that the resultant velocity is the hypotenuse, which means that 4^2+3^2=v^2, where v=5 mi/h. And that's only part of the answer. The whole answer in the textbook says 5 mi/h at an angle of ≈53.13° to the bank. But how do I find the angle to the bank?

Have you drawn a diagram?
 
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Well, the problem didn't mention as perpendicular to the bank. So that means the answer is just 5 mi/h, right?
 
Math9999 said:
Well, the problem didn't mention as perpendicular to the bank. So that means the answer is just 5 mi/h, right?

A velocity always has a direction.
 
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So how to figure the angle out?
 
Math9999 said:
So how to figure the angle out?
By the assumption I've suggested and the diagram @PeroK has suggested. Figure out the remaining angles of your triangle.
 
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cos-1(3/5)≈53.13°, am I right?
 
Math9999 said:
cos-1(3/5)≈53.13°, am I right?

I suspect your calculator can tell you that!
 
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Okay, thank you!
 

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