You are riding on a jet ski at an angle upstream on a river flowing with a speed of 2.8 m/s. Suppose the jet ski is moving at a speed of 18 m/s relative to the water.
(a) At what angle must you point the jet ski if your velocity relative to the ground is to be perpendicular to the shore of the river?
(b) If you increase the speed of the jet ski relative to the water, does the angle in part (a) increase, decrease, or stay the same?
If the velocity is perpendicular, wouldn't the angle be 90degrees? That seems too easy; any help?
The upstream velocity is 18sinTheta. That has to equal the stream velocity, right? Solve for Theta
Well, well, well! Looks like you've got a fun little physics problem on your hands. Let's dive right in!
(a) So, you want your velocity relative to the ground to be perpendicular to the shore of the river. This means that the horizontal component of your velocity (due to the river's flow) should cancel out the horizontal component of your jet ski's velocity relative to the water.
Let's call the angle between the jet ski's velocity relative to the water (18 m/s) and the shore of the river θ. The horizontal component of the jet ski's velocity relative to the water is given by 18 m/s * cos(θ).
The river's flow is 2.8 m/s, and since we're going upstream, its velocity is in the opposite direction. The horizontal component of the river's flow is 2.8 m/s * cos(180°) = -2.8 m/s.
For the net horizontal velocity to be zero, the horizontal components of the jet ski's velocity relative to the water and the river's flow must cancel each other out. So we have the equation:
18 m/s * cos(θ) + (-2.8 m/s) = 0
Solving this equation will give us the value of θ, the angle at which you must point the jet ski. Can you give it a go?
(b) Ah, the good ol' speed increase conundrum! If you increase the speed of the jet ski relative to the water, the angle in part (a) will actually stay the same.
Why, you ask? Well, the river's flow doesn't change, and the horizontal component of the jet ski's velocity relative to the water still needs to cancel out the horizontal component of the river's flow to keep your velocity perpendicular to the shore.
So, go ahead and solve for θ in part (a) using the given values, and you'll have your answer! Remember, even though the angle may seem easy, it's always fun to put some physics behind it. Happy calculating, my friend!
To find the angle at which you must point the jet ski to have a velocity perpendicular to the shore, we can use the concept of vector addition.
(a) Let's break down the velocities into their respective components:
Velocity of the jet ski relative to the ground: v_g = 18 m/s
Velocity of the river stream: v_r = 2.8 m/s
We want the velocity relative to the ground to be perpendicular to the shore. This means that the horizontal component of the river stream velocity should cancel out the horizontal component of the jet ski velocity.
The horizontal component of the jet ski velocity relative to the water is given by:
v_x = v_j*cos(θ)
The horizontal component of the river stream velocity is v_r*cos(180°) = -v_r (since it acts in the opposite direction).
For the velocities to cancel out in the horizontal direction, we have:
-v_r = v_j*cos(θ)
Simplifying this equation, we find:
v_j*cos(θ) = -v_r
18*cos(θ) = -2.8
Now, solve for the angle θ:
θ = arccos(-2.8/18)
θ ≈ 98.7°
So, you must point the jet ski at an angle of approximately 98.7 degrees upstream to have a velocity relative to the ground perpendicular to the shore.
(b) If you increase the speed of the jet ski relative to the water, the angle in part (a) will stay the same. The angle is determined by the ratio of the horizontal component of the jet ski velocity to the river stream velocity. As long as this ratio remains constant, the angle will not change.