Frames of Reference and Relative Velocity

[whoareyou]

A swimmer who achieves a speed of 0.75 m/s in still water swims directly across a river 72 m wide. The swimmer lands on the far shore at a position 54 m downstream from the starting point.
(a) Determine the speed of the river current.
(b) Determine the swimmer?s velocity relative to the shore.
(c) Determine the direction the swimmer would have to aim to land directly across from the starting position.

I am having trouble understanding the actual physics of this question. I mean I get the math: a) 72/0.75 = 96s --> 54/96 = 0.56m/s but I don't understand why. If the swimmer is swimming with the current, should it the speed be faster? And are my calculations even right because it says he swims across the river, but is then 54m downstream. So there are two different directions ...

Can anybody explain the physics of this question to me please?

 

[cepheid]

A swimmer who achieves a speed of 0.75 m/s in still water swims directly across a river 72 m wide. The swimmer lands on the far shore at a position 54 m downstream from the starting point.
(a) Determine the speed of the river current.
(b) Determine the swimmer?s velocity relative to the shore.
(c) Determine the direction the swimmer would have to aim to land directly across from the starting position.

I am having trouble understanding the actual physics of this question. I mean I get the math: a) 72/0.75 = 96s --> 54/96 = 0.56m/s but I don't understand why. If the swimmer is swimming with the current, should it the speed be faster? And are my calculations even right because it says he swims across the river, but is then 54m downstream. So there are two different directions ...

Can anybody explain the physics of this question to me please?

Let's say that the river is oriented with the vertical axis in our coordinate system (i.e. vertical = parallel to the river and horizontal = perpendicular to the river). So the swimmer is swimming at 0.75 m/s horizontally. Meanwhile, the current is carrying him at some unknown speed downstream (i.e. vertically). So, his/her velocity relative to the bank is the vector sum of these two (meaning that it points somewhere in between entirely horizontal and entirely vertical). As a result, at the end of the trip, the swimmer will have moved BOTH horizontally (across to the opposite bank) and vertically (downstream). Indeed, that is the case. The swimmer ends up on the opposite shoreline (72 m horizontally from where he was before) but he is 54 m farther along that shoreline compared to when he started. So, relative to the shore, he actually moved in a diagonal line that is the hypotenuse of the right triangle formed by his horizontal and vertical motions (i.e. 54 m is one side of the triangle, 72 m is the perpendicular side, and the third side, the hypotenuse, is his actual net displacement). Does that help?

 

[PeterO]

A swimmer who achieves a speed of 0.75 m/s in still water swims directly across a river 72 m wide. The swimmer lands on the far shore at a position 54 m downstream from the starting point.
(a) Determine the speed of the river current.
(b) Determine the swimmer?s velocity relative to the shore.
(c) Determine the direction the swimmer would have to aim to land directly across from the starting position.

I am having trouble understanding the actual physics of this question. I mean I get the math: a) 72/0.75 = 96s --> 54/96 = 0.56m/s but I don't understand why. If the swimmer is swimming with the current, should it the speed be faster? And are my calculations even right because it says he swims across the river, but is then 54m downstream. So there are two different directions ...

Can anybody explain the physics of this question to me please?

Perhaps a similar situation will make it clearer:

A conveyor belt runs from a mine to a factory.
An ant walks across a conveyor belt, 30 cm wide, at a speed of 3mm per second. When it reaches the other side the ant is 5 m closer to the factory than when the ant started.
a) how fast is the belt running etc etc

( in this case the ant is simply not fast enough to get straight across the conveyor mechanism)

Now the swimmer
For the first part.
As the swimmer is about to begin, he aims at a strip of water, say 2m wide, stretching across the river. he swims along that strip. While he is swimming, that strip, and all the rest of the river, moves downstream, so that when he has got to the other side of the strip of water, the strip of water is 54m further down-stream than when he started.

For the second part,
As the swimmer is about to start, he aims at a strip of water angled up-stream [given your figures, at about 45 degrees??]. He again swims along that strip, but the strip will have moved downstream just enough that when he reaches the end of the strip, it is positioned exactly opposite the point where he began to swim.

 

[NascentOxygen]

I used to have a lot of trouble with these at school, too. I suggest you commit to memory this single formula and it will serve you well for the rest of your life:

V_a = V_{a rel b} + V_b

where V_{a rel b} is read as "velocity of a relative to b".

If this means nothing much to you at the moment, perhaps convert it to money, or weight, or something that is meaningful to you, so that you can work it out and write it down every time you need it. Example, if Fred weighs 25kg more than you, and you weigh 70kg, how much does Fred weigh? Here's the equation:

W_f = W_{f rel u} + W_u

For scalar quantities, these add as scalars. For vector quantities, these add as vectors and you need to draw a vector triangle to perform that vector addition.

Now, the swimmer makes progress through the water at his swimming speed, this is V_{s rel w}. To find his progress as far as the fixed reference point of someone on the bank is concerned, we need:
V_s = V_{s rel w} + V_w

Progress (i.e., distance) is directly proportional to speed, so as vector quantities, both displacement and velocity diagrams form congruent (or similar) triangles.

 

[rwisz]

Sure, I'd be happy to help explain the physics behind this question. Let's break it down into smaller parts and then put it all together.

First, let's talk about frames of reference. A frame of reference is a system of coordinates that we use to describe the motion of an object. In this case, we can use two frames of reference: one for the swimmer and one for the river. The swimmer's frame of reference is moving with the swimmer, while the river's frame of reference is stationary.

Now, let's talk about relative velocity. Relative velocity is the velocity of an object with respect to another object. In this case, we are interested in the swimmer's velocity relative to the shore. This means we are looking at the swimmer's velocity with respect to the stationary frame of reference of the river.

Now, let's look at the situation described in the question. The swimmer is swimming with a speed of 0.75 m/s in still water. This means that in the river's frame of reference, the swimmer's velocity is 0.75 m/s. However, the swimmer is also moving with the river current, which has a velocity of v (we'll determine this value later). This means that in the swimmer's frame of reference, the river's velocity is -v (since it is moving in the opposite direction of the swimmer's motion).

Now, let's look at the swimmer's motion across the river. The swimmer is swimming directly across the river, which has a width of 72 m. This means that the swimmer will take 72/0.75 = 96 seconds to cross the river. During this time, the swimmer is also being carried downstream by the river current. The distance the swimmer travels downstream is 54 m.

To determine the speed of the river current, we can use the formula v = d/t, where v is the velocity, d is the distance, and t is the time. Using the values we have, we can calculate the river current's velocity as 54 m/96 s = 0.56 m/s.

Now, to determine the swimmer's velocity relative to the shore, we need to take into account the swimmer's velocity in the river's frame of reference (0.75 m/s) and the river current's velocity in the swimmer's frame of reference (-0.56 m/s