Powerboat moving at an angle across a river with current

[chenying]

Homework Statement

A 210-m-wide river has a uniform flow speed of 2.4 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 8.4 m/s with respect to the water. There is a clearing on the north bank 17 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

Homework Equations

Superimpose an xy coordinate system, with the water moving in the positive x direction. The time required to cross the river width is the same as the time to move upstream to the clearing. Thus the boat's velocity bg relative to the shore must have a negative x component.

The Attempt at a Solution

Find the x and y components of the vector. Use arctan to find the angle. arctan(210/17) =85.37186
2.4t - 8.4cos (theta)t = 17
8.4sin(theta)t = 210
solve for t to find theta.
I've tried a ton of methods but I'm not getting the right answer. Please help.

 

[anathema]

Hi there,

Thank you for your question. It seems like you are on the right track with your solution attempt, but there are a few issues that may be causing you to not get the correct answer.

Firstly, when using the equation 2.4t - 8.4cos (theta)t = 17, it is important to note that the units of velocity (m/s) and time (s) must match. In this case, the velocity of the river is given in m/s, but the velocity of the boat is given in m/s with respect to the water. This means that the boat's velocity relative to the shore will actually be 2.4 m/s + 8.4 m/s = 10.8 m/s. So the equation should be 10.8t - 8.4cos(theta)t = 17.

Secondly, when using the equation 8.4sin(theta)t = 210, it is important to note that the 210 m is the distance from the south bank to the north bank, not just the distance from the boat's starting position to the north bank. This means that the equation should be 8.4sin(theta)t = 210 - 17 = 193.

Lastly, when solving for t, you should use the quadratic formula since the equation is quadratic. This will give you two solutions for t, but only one of them will be physically meaningful (the other one will be negative). Once you have found the correct value for t, you can use it to solve for theta using the equation 10.8t - 8.4cos(theta)t = 17.

I hope this helps! Let me know if you have any further questions. Keep up the good work!

 

[tomcue]

I would approach this problem by breaking it down into smaller parts and using known equations and principles to solve it.

First, I would draw a diagram of the situation and label all the given information, such as the river width, flow speed, boat speed, and the location of the two clearings. This will help visualize the problem and identify any unknowns.

Next, I would use the principle of relative motion to determine the velocity of the boat relative to the shore. We know that the boat has a constant speed of 8.4 m/s with respect to the water, which is moving at 2.4 m/s in the positive x direction. Therefore, the velocity of the boat relative to the shore is (8.4 m/s, -2.4 m/s) or (8.4 m/s, -2.4 m/s) in vector notation.

Now, we can use the equation v = d/t to solve for the time it takes for the boat to cross the river. We know that the distance the boat must travel is the river width (210 m) and the velocity is the boat's velocity relative to the shore. Therefore, we can write the equation as 210 m = (8.4 m/s, -2.4 m/s) x t. Solving for t, we get t = 21.875 seconds.

To find the angle at which the boat must be pointed, we can use the trigonometric relationship tan(theta) = opposite/adjacent. In this case, the opposite side is the y component of the boat's velocity relative to the shore (which is -2.4 m/s) and the adjacent side is the x component (which is 8.4 m/s). Therefore, tan(theta) = (-2.4 m/s)/(8.4 m/s) = -0.2857. Taking the inverse tangent, we get theta = -15.5 degrees.

Finally, to find the time it takes for the boat to reach the clearing on the north bank, we can use the equation d = vt, where d is the distance from the south bank clearing to the north bank clearing (17 m) and v is the boat's speed relative to the shore (8.4 m/s). Therefore, t = 17 m/8.4 m/s = 2.02 seconds.

In conclusion, the boat must be pointed at an angle of -15