You're visiting the shark tank at the aquarium when you see a 2.5...

Question

You're visiting the shark tank at the aquarium when you see a 2.5-m-long shark that appears to be swimming straight toward you at 2.0 m/s. What is the shark's actual speed through the water? You can ignore the glass wall of the tank.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a submarine exhibit. A 0.8 m submarine prototype is witnessed moving into a vast water tank at a speed of 1.3 m per second. As perceived by an observer, the observer sees the submarine coming directly toward him from behind the tank's glass wall, determine the submarine's true speed in the water, assume that the thickness of the glass does not affect the perceived motion of the submarine. So our end goal ultimately for this particular problem is we're asked to figure out what the submarine's true speed is in the water. Awesome. And that's what we're f that's what our final answer is that we're ultimately trying to solve for is we're trying to figure out the submarine's true speed. Awesome. So with that in mind, let's look at our multiple choice answers to see which answer might be our final answer. Let's also note that they're all in the same units of meters per second, which is the standard unit for speed. So let's read them off. So A is 0.8 B is 1.3 C is 1.7 and D is 2.3. Awesome. So first off, let us assume that the wall of the tank represents the boundary between the water and the air. Let us also ignore the effect of glass for this particular problem. So now, first off, let us create a diagram of this problem to better visualize what is happening before we go any further with the solving process. So I've gone ahead and put together a fancy diagram of what's going on in this particular problem. So the blue square rectangular shape represents the tank. And we have this black dot on the far left which represents the submarine. And then moving towards the right side of our picture, we have the another black dot representing the submarine's image and cutting through the center of our submarine dot And our image dot is the optical axis. So the optical axis cuts it in half. And we can see that from the image dot to the boundary at the very end of our glass tank is the distance S and from our submarine dot to the glass wall on the other side is our s prime distance. And we can see how the light is reflected reflected off of the submarine and then will strike the wall of the tank and then shoot out. And that's what this thin, this very thin like arrow line is denoting the light's travel path. So it will strike the glass boundary and then it will the light will bend in another direction. OK. So now that we have an idea of what's happening visually in this problem, let us use our diagram. And when we use our diagram, we will note that the relation of the image formation by the refraction at the plane surface can be written as. So using our diagram, we can write the following equation, let's call it equation one. So S prime is equal to negative N two divided by N one multiplied by S where S prime is the distance of the image formed from the surface, which is from the observer's perspective. After refraction occurs, N one represents the refractive index of medium where the incident ray is coming from which in this case, it is water. So N one equals water. And N two is the refractive index of medium where the refracted ray is traveling through which in this case, it's air. So let's make a quick little note here because that's a lot of information. So let's note really quick that N one is water. So the refractive index of water and N two is the refractive index of air. So that's a quick little note. So moving right along S is equal to the distance of the object from the surface. So note that the minus sign in our equation one. What is the reason why there is a minus sine disease is, it tells us that there's an that this is in virtual image. So a virtual image will be produced. That's why there's a negative sign in our equation one. Also let us note that the submarine is moving. Therefore S prime and S distances will change over time. So the refractive indexes N one and N two will be assumed to be constant. And we can derive equation one with respect to time to write that Ds prime. So Ds prime of DT is, we're taking equation one and we're deriving it with respect to time is equal to negative N one divided by N two DS of DT. So note that we can write that the speed of the submarine as seen by the observer as. So we can write that as B prime is equal to DS prime of DT. And we can write the speed of the submarine in the water as V which is equal to DS of DT or DS by DT, I should say. Thus, we could write that V prime is equal to the absolute value of negative N two divided by N one multiplied by V which we can simplify and write that V is equal to N one divided by N two multiplied by V prime, which is equal to 1.33 which is the refractive index of glass divided by 1.00 which is the refractive index of water. And we need to multiply this by the speed which we're told it's moving at 1.3 m per second. And to be specific, that's the speed of our submarine moving in the tank. So when we plug that into our calculator, we will find that the speed or the true speed I should say is equal to 1.7 m per second. And that is our final answer when we round to one decimal place. So once again to repeat, the submarine's true speed in the water is 1.7 m per second. And this corresponds with multiple choice answer letter C which once again, it's 1.7 m per second. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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You're visiting the shark tank at the aquarium when you see a 2.5-m-long shark that appears to be swimming straight toward you at 2.0 m/s. What is the shark's actual speed through the water? You can ignore the glass wall of the tank.

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