YT YouTube2Text

YouTube2Text → Transcripts → 2.25 ভেক্টর relative velocity math problems | আপেক্ষিক বেগের বৃষ্টি ছাতার সমস্যা- part -03|

2.25 ভেক্টর relative velocity math problems | আপেক্ষিক বেগের বৃষ্টি ছাতার সমস্যা- part -03| — Transcript

by Apurbo physics · 6,288 words · 905 segments · language en · Watch on YouTube

Open in transcriber Summarize with AI Download SRT/VTT

Full transcript

0:10In the last video, we did a few math
0:12problems related to relative velocity.
0:14In today's video, we will look at a few
0:16more problems on relative velocity. If
0:19I categorize the relative velocity
0:21problems that you find difficult or
0:23troublesome into two parts, the first
0:25category would be umbrella-holding
0:26problems in rain, and the second
0:28category would be car windshield
0:30wetting problems in rain. Believe me,
0:33these problems are very easy. But many
0:36students find these problems quite
0:38difficult. One reason for this
0:40difficulty is similar to trying to
0:42learn how to form sentences in English
0:44without first learning the alphabet;
0:45it's going to feel very difficult.
0:49Similarly, if you try to jump into
0:51complex math without mastering the
0:53small, basic problems, this simple
0:55topic will seem very hard to you.
0:58Another reason is the lack of "feel"
1:00for the subject. Meaning, if you are
1:03only doing it on paper and cannot feel
1:05the situation realistically using
1:06common sense, it will seem difficult.
1:09So, the first step to making these math
1:11problems easy is to bring in that
1:13feeling. That is, we need to examine
1:16whether the math we are doing actually
1:18holds true in real life, and another
1:19task will be to gradually move from
1:21small, basic problems toward the
1:23difficult ones. So, if you watch the
1:27math problems I solve in this video in
1:29order and with full attention, I hope
1:31you won't have any more trouble with
1:32relative velocity problems. Okay, let's
1:35begin. So, first, let's look at the
1:37umbrella-holding problems in the rain.
1:40I have already solved one
1:41umbrella-holding problem for you before
1:43. So, now we will solve many such
1:45problems. Previously, we saw a boy just
1:47running in the rain. Meaning, there was
1:49the passenger's velocity and the rain's
1:51velocity. Now, we will look at cases
1:53where there is passenger velocity, rain
1:55velocity, and also wind velocity. So,
1:57let's start with a very simple math
1:59problem. Our first problem states that
2:03rain is falling vertically at 5 meters
2:05per second. In the first case, it is
2:10mentioned that there is no wind. No
2:13wind. Meaning, we will not include wind
2:17in our calculations here. Our first
2:20problem asks how a stationary passenger
2:22should hold an umbrella to stay
2:23protected from the rain. The second
2:28problem asks how the passenger should
2:29hold the umbrella if they start walking
2:31at 10 meters per second. The third one
2:36asks if the angle of the umbrella with
2:38the vertical increases as the
2:39passenger's velocity increases. Explain
2:43this mathematically. So let's get
2:45started. Our first task will be to draw
2:47the diagram. Rain is falling vertically
2:50at 5 meters per second. So, this is the
2:53rain. We have drawn this rain.
2:57Represented by Vr. The velocity is 5
2:59meters per second. Now look, this is
3:02the flat ground. A passenger is
3:05standing here. The passenger is
3:07standing like this. So the passenger is
3:10standing still. The passenger has no
3:12velocity. How is the rain falling?
3:14Vertically. The question asks how a
3:17passenger standing still should hold an
3:18umbrella. Is there any need to do math
3:20for this? You are standing still. Rain
3:23is falling vertically onto your head.
3:25So, if you have an umbrella, to protect
3:26yourself from the rain, you will hold
3:28the umbrella vertically right over your
3:30head. There is no need for any math at
3:32all. So why did I give this? There is a
3:34reason why I gave it. We will
3:35understand in a little while. So if we
3:37write the answer in this case, what
3:39will we write? Hold the umbrella
3:40vertically. Number one is done. Hold
3:44the umbrella vertically. In this case,
3:50we don't even need to do any
3:51calculation. Now let's look at the
3:54second case. What does the second one
3:56say? How should the umbrella be held if
3:58the passenger starts walking at 10
4:00meters per second? First, I told you to
4:03get a feel for it and think about the
4:04real-world scenario. You were standing
4:07still and you held the umbrella
4:08vertically over your head. Now the rain
4:10is falling. You have started walking in
4:12this direction. So, as soon as you
4:13start walking or running forward like
4:15this. Notice that in real life, we tilt
4:19the umbrella at an angle with the
4:20vertical. That is, the umbrella is
4:23tilted in the direction we are running.
4:26So let's see what happens. I told you
4:30that the velocity of the rain is the
4:31relative velocity of the rain with
4:33respect to the passenger. That is, when
4:35the passenger moves forward, the
4:36velocity of the rain relative to the
4:38passenger will seem to be coming from
4:40this direction. So, for relative
4:41velocity, we said we have to reverse
4:43the velocity of the reference point.
4:44Since it is the velocity of the rain
4:46relative to the passenger, we will
4:48reverse the passenger's velocity. So
4:49notice that the passenger was moving
4:51this way with a velocity of 10 meters
4:53per second, this is the passenger's
4:55velocity. Since the rain's velocity is
4:57relative to the passenger, we reverse
4:59the passenger's velocity here,
5:01considering the raindrop. Meaning, we
5:03will not keep it this way anymore. So,
5:05the passenger's velocity in this
5:06direction will be what? 10 meters per
5:08second. How is the rain falling? The
5:09rain is falling vertically. At what
5:11velocity? At a velocity of 5 meters per
5:13second. So the first task was to
5:15reverse, and the second task was to
5:16find the resultant after reversing. So
5:18look, the reversal task is done. We
5:20have done such math before. So this
5:22velocity vector is no longer on this
5:24side. After reversing, the resultant of
5:26these two will come along this line.
5:28The resultant will come along this line
5:29. So the rain will hit the passenger
5:32here from this direction. So if the
5:36passenger wants to protect themselves
5:37from the rain by holding an umbrella,
5:39they must hold the umbrella with the
5:40vertical. Look, with the vertical line,
5:44they have to hold the umbrella tilted
5:46at an angle theta. This means to
5:48protect themselves from the rain, they
5:50will now hold the umbrella. We have
5:51been asked to find the value of this
5:53angle theta for holding the umbrella.
5:54Look, how will they hold the umbrella?
5:56So, I was saying the angle to hold the
5:58umbrella with the vertical is theta.
5:59Now, let's find theta in a very short
6:01way. Can we write down tan theta? Let's
6:04look at tan theta carefully. This angle
6:06is theta. This is 10, and what was this
6:09? This was 5. So, if we shift this up.
6:12If I bring it to this side, this will
6:14also be 5. So, tan theta equals
6:17perpendicular by base, which is 10 by
6:195. Therefore, theta equals tan inverse
6:222, and calculating this gives 63.43.
6:27Degrees; meaning, to stay protected
6:29from the rain, one must hold the
6:31umbrella at a 63.43-degree angle with
6:33the vertical. I hope we understood this
6:35concept. Now look at what number three
6:38says: analyze mathematically if the
6:40angle of the umbrella with the vertical
6:42increases when the pedestrian increases
6:43their speed. It would be great if I had
6:45an umbrella; camera, do we have the
6:47umbrella? Let's see, look, this is it.
6:51When rain falls vertically, I hold the
6:53umbrella straight above my head like
6:55this. This was the first case,
6:57vertically; now, if I start walking
6:59this way, notice, I am moving in this
7:01direction. So when I move this way, I
7:04tilt the umbrella like this. At what
7:06angle with the vertical do I tilt it? I
7:08tilt it at this 63.43-degree angle.
7:11Think about it for a moment. If I
7:13increase my speed in the direction I am
7:15walking—that is, if I go faster than
7:17before—wouldn't I tilt the umbrella
7:19even more like this? So, you see, the
7:22umbrella has to be tilted at a larger
7:24angle with the vertical line. That is
7:27what the problem asks: if the
7:28pedestrian increases speed, does the
7:30umbrella angle with the vertical
7:32increase? Explain and analyze this
7:33mathematically. So, let's see if we can
7:35explain and analyze it mathematically.
7:38Suppose for the third case, imagine our
7:40speed was 10, and now we make the speed
7:4215. So if the speed is 15, then after
7:44reversing it, the velocity on this side
7:47will be 15. It will be on this side.
7:49And the rain falling is at a speed of 5
7:51meters per second. Then, the resultant
7:53of these two will strike the pedestrian
7:55like this. So, if the pedestrian wants
7:58to be protected from the rain, they
8:00must now hold the umbrella with the
8:01vertical. The umbrella must be held at
8:04an angle of theta. This angle is that
8:05theta angle. So, what was our target?
8:08To find this theta angle. Look, if this
8:10is 5, then this is 5. So if we
8:11calculate this angle, let's write it at
8:13once: theta equals tan inverse
8:15perpendicular 15, divided by base 5,
8:17that is, tan inverse 3. Calculate it,
8:19and it will come to 71.56 degrees. So,
8:22now to be protected from the rain, one
8:25must tilt the umbrella at a larger
8:27angle of 71.56 degrees with the
8:28vertical. Earlier, when the speed was
8:3110, at what angle did the umbrella need
8:33to be held? At a 63.43-degree angle. So
8:37, you see, the more he increases his
8:39speed, the more he has to tilt the
8:40umbrella at a greater angle from the
8:42vertical to stay protected from the
8:44rain. I hope you have understood this
8:47concept. Try to feel this thing in a
8:49practical sense. Then you will be able
8:51to understand it even more easily. Now,
8:53we are going to do the same type of
8:55math. But notice, earlier there was no
8:57wind. So, what are we going to do now?
9:00We will add wind to it and then see how
9:01he needs to hold the umbrella. The
9:06second one is vertical rain at 5 meters
9:08per second and wind blowing from east
9:10to west at 2 meters per second. Notice
9:15earlier there was no wind. Now I have
9:17added wind. And in which direction is
9:19the wind blowing? It is blowing from
9:20the east to the west. With what
9:22velocity? With a velocity of 2 meters
9:23per second. And the rain is falling the
9:25way it was falling before. Vertical
9:26rain at 5 meters per second. Now, the
9:30first math problem asks how a
9:31stationary traveler should hold the
9:33umbrella. So, let's see how a
9:35stationary traveler should hold the
9:37umbrella. Let's draw it. First, let's
9:39draw the rain. The rain is falling
9:41vertically with what velocity? With a
9:43velocity of 5 meters per second. And
9:45what else is there? Wind. In which
9:47direction? From east to west. Let's say
9:49this direction is the west direction.
9:50We call this direction the east
9:52direction. So, the wind is blowing from
9:55east to west at 2 meters per second.
9:58Now, in which direction should a
10:00stationary observer hold the umbrella
10:02to stay protected from the rain? That
10:05is our question. Think about it for a
10:06moment first. Imagine the rain is
10:08falling vertically. There is no wind.
10:11In that case, how would a stationary
10:13observer hold the umbrella? Since the
10:16rain is falling vertically, he would
10:18hold the umbrella vertically over his
10:20head. Which is what we did in the
10:22previous math problem. Now the question
10:25is, if there is vertical rain and a
10:27wind comes from this side, can the
10:29traveler still hold the umbrella
10:30vertically? No, he cannot. Because the
10:35wind will hit the raindrops that are
10:37falling vertically. Then the raindrops
10:40will no longer be able to fall along
10:42that vertical path. That is, the path
10:45of the rain will deflect, and it will
10:47fall along the resultant velocity of
10:48the wind and the rain. So, notice that
10:52if a traveler is standing still, he
10:54will no longer hold the umbrella
10:55vertically over his head. He will then
10:59definitely hold the umbrella at a theta
11:01angle from the vertical by tilting it.
11:04Think about it yourself; imagine you
11:06are standing still in one place. The
11:08rain is falling vertically. You are
11:10holding the umbrella vertically. But if
11:12suddenly you see a gust of wind coming
11:13from that direction. Will the rain
11:15still fall on your head vertically then
11:17? Then, if you have an umbrella in your
11:20hand, you will certainly tilt it in the
11:21direction from which the gust of wind
11:23is coming. Are you getting the feel of
11:25it? So we saw that for a stationary
11:27observer, if the rain falls vertically,
11:29they will hold the umbrella vertically
11:31over their head. But if there is wind,
11:34you must tilt the umbrella in the
11:36direction from which the wind is
11:37blowing. So, this thing can be felt,
11:39right? It can be understood and
11:41explained, can't it? Okay. Now we need
11:43to calculate. At what angle should the
11:45umbrella be held? For a stationary
11:47observer, these raindrops will fall
11:49with a resultant velocity of the rain's
11:51velocity and the wind's velocity. Then
11:54this passenger must hold the umbrella
11:56along this direction. So, at what angle
11:58with the vertical? At an angle of theta
12:00(θ). Our target is to find this theta
12:02angle. So similarly, if this is five,
12:03then along this direction it is five
12:05and this one is two. So if we calculate
12:08the value of the theta angle, it is tan
12:10inverse of 2/5 with the vertical. Then
12:13the value of this angle will be 21.80
12:15degrees. Meaning, they must hold the
12:18umbrella at this angle with the
12:19vertical. Note that in this case, we
12:21did not reverse anyone's velocity. We
12:23did not calculate the relative velocity
12:24of one with respect to another. Where
12:26we first reverse the velocity of the
12:27one with respect to whom the velocity
12:29is being calculated. And then find the
12:30resultant after reversing it. We did
12:32not do this task this time. Now, there
12:34is something for you to think about
12:35here. In previous math problems, when
12:37it was the velocity of rain with
12:38respect to a boy or a passenger. We did
12:40reverse the passenger's velocity. But
12:42in this case, when dealing with the
12:44velocity of wind and rain, notice we
12:46did not reverse any velocity. We
12:47directly calculated the resultant of
12:49the wind velocity and rain velocity,
12:51and the rain will fall with this
12:53resultant velocity of wind and rain. So
12:55from this, one thing can be said: if it
12:57is about a boy and rain velocity, then
12:59for rain velocity with respect to the
13:01boy, you must reverse the boy's
13:03velocity to find the resultant relative
13:05velocity. But if it is like this, where
13:07there is rain velocity and wind
13:08velocity, then they do not have a
13:10relative velocity. Instead, they have a
13:12resultant. That is, the raindrop is
13:14falling with the rain's velocity. What
13:17is pushing this raindrop? The wind.
13:19Then this raindrop falls with the
13:21resultant of these two velocities. So
13:23look, in this case, it is not relative
13:25velocity. In this case, it is the
13:26resultant. So let me write it down like
13:28this. You need to keep this very
13:30clearly in mind. First case: passenger
13:35and rain; if there is a passenger and
13:38rain, then it is relative velocity.
13:43What needs to be done? The passenger's
13:45velocity must be reversed to find the
13:47resultant. For example, this is the
13:59rain velocity. And this is the
14:03passenger's velocity. Since it is the
14:06rain velocity relative to the passenger
14:08, we will reverse the passenger's
14:10velocity. After reversing, we will
14:11calculate the resultant. This is what
14:13we will do in the case of rain and a
14:15passenger. The second one is rain and
14:19wind. If there is rain and wind, there
14:24is no relative velocity here. So in
14:26this case, it is a direct resultant.
14:32The raindrop is falling with the rain's
14:34velocity. And the wind comes and pushes
14:37this velocity of the rain. It pushes
14:39the rain, slanting it, and it falls
14:40with a resultant velocity. So, it is a
14:43direct resultant. Okay? It is like the
14:46river-boat scenario. Look, the boat's
14:48velocity acts on the boat. And the
14:50current comes and pushes what? The boat
14:51. Then what happens? It takes the
14:53resultant of the current's velocity and
14:55the boat's velocity. The boat moves
14:56with that resultant velocity. It is the
14:58same here. The rain will fall with the
15:00resultant of the wind velocity and rain
15:01velocity. Okay? Another one, the third
15:04case, is the wind and the traveler. If
15:10we think about the wind and the
15:11traveler, what will happen? Then it
15:13will be the relative velocity of the
15:14wind with respect to the traveler. So,
15:16in this case, it will also be relative
15:18velocity. Keep these three things in
15:20mind. Then all the relative velocity
15:22math problems will become a piece of
15:23cake. So, let's look at an example here
15:25. Suppose the traveler is going this
15:27way. This is the traveler's velocity,
15:29and the wind is going this way. This is
15:31the wind velocity. So look, pay close
15:33attention. Let me state the values.
15:35Suppose the traveler's velocity is 5
15:37meters per second. And the wind's
15:39velocity is 2 meters per second. So,
15:40pay close attention. When this traveler
15:43goes this way, it will seem to the
15:45traveler that the wind velocity is much
15:47higher. Look closely at the car example
15:50I mentioned. That I am driving a car
15:52this way. Someone is coming from the
15:54opposite direction in a car. Suppose my
15:55car's velocity is 5 meters per second.
15:58And someone is coming from the opposite
15:59direction. That car's velocity is 2
16:01meters per second. Then wouldn't that
16:03car's velocity seem higher to me? How
16:05much would it seem? 5 plus 2, which is
16:087 meters per second. So what is this?
16:10That car's velocity with respect to me.
16:12It is exactly the same situation. This
16:14traveler is moving this way at a
16:15velocity of 5 meters per second. The
16:17wind is coming from the opposite
16:18direction. So, it will seem to this
16:20traveler that the wind is coming with
16:21even more velocity. Because they are
16:23running towards the wind's velocity. So
16:25, the wind's velocity will seem to them
16:27to be 5 plus 2, which is 7 meters per
16:29second. So, 7 meters per second is
16:31whose velocity with respect to whom?
16:33Wind's velocity with respect to the
16:35traveler. So look, we reversed it in
16:42this case. We took the resultant after
16:43reversing. With respect to whom? With
16:45respect to the traveler. So, reverse
16:46the traveler's velocity. In the
16:48opposite direction of where it was, how
16:49much is that? Five. The wind's velocity
16:51was this way, how much is that? Two.
16:52Adding both together, how much does it
16:53ultimately become? Seven. So look, we
16:55are finding the resultant after
16:57reversing. So, let me say it again.
16:58Traveler and rain will be relative
17:00velocity. Resultant after reversing.
17:02Wind and traveler will be relative
17:03velocity. Resultant after reversing.
17:05But if we think about rain and wind, it
17:07is a direct resultant. This is not
17:09relative velocity. If you can keep
17:11these three cases in mind, you won't
17:13have any more trouble with relative
17:15velocity math. We will look at a few
17:17more categories of this math. Then we
17:19will be able to understand it. The
17:20second problem of this stimulus. Look,
17:24the second one says, if the pedestrian
17:26starts walking from west to east at 6 m
17:28/s, in which direction will they hold
17:30the umbrella? That is, the pedestrian
17:33has been made dynamic this time.
17:35Earlier, the pedestrian was stationary.
17:37Now, the pedestrian has started walking
17:39in this direction. So, let's draw it
17:42here. This is the pedestrian. So, the
17:44pedestrian is moving in this direction
17:46with what velocity? With a velocity of
17:496 m/s. How is the rain falling? The
17:52rain is falling vertically. With what
17:54velocity? With a velocity of 5 m/s.
17:56What else is there? This is the west
17:58direction. This is the east direction.
18:00So, the wind is blowing from the east
18:03to the west at a velocity of 2 m/s. So,
18:06look, in this case, there are three
18:08things. There is rain, wind, and the
18:12velocity of the pedestrian. When there
18:16are three such things, it is seen that
18:18a lot of confusion starts in the head.
18:20There is no reason for confusion to
18:22start. There are three, right? Work on
18:24two first. If you work on two first,
18:27then instead of two, there will be one.
18:29Then work on the rest with that one. So
18:31, which two can be worked on first?
18:34Very easy. We will calculate with the
18:37two that are easier to calculate with
18:38first. Which two would be easier to
18:41take first? Take the two that are on
18:43the same straight line first. Look, the
18:46wind and the pedestrian are on the same
18:48straight line. They are parallel to
18:51each other. Even if the direction is
18:53different, they are parallel to each
18:55other. For this reason, calculate these
18:57two first. So look, let's calculate the
18:59wind relative to the pedestrian first.
19:03So we write here, wind relative to the
19:05pedestrian. There is no need to
19:08calculate the rain first. Relative to
19:10the pedestrian, how does the wind seem
19:12to you? The pedestrian is going in this
19:13direction with what velocity? With a
19:15velocity of 6 m/s. The wind is coming
19:17from the opposite direction with what
19:19velocity? With a velocity of 2 m/s. So,
19:21to you, the pedestrian, how will the
19:23wind's velocity seem? If the wind comes
19:26from this side at 2 m/s, won't the
19:28wind's velocity seem higher to you? And
19:31how much will that be? 6 plus 2. If
19:33it's hard to understand this way, then
19:35think of yourself running, and instead
19:37of wind, a friend is running towards
19:39you. Then wouldn't you feel that the
19:41friend is running towards you even
19:42faster, and at what velocity? 6 plus 2
19:45equals 8 m/s. So we can write, the wind
19:48relative to the pedestrian will be 6
19:51plus 2 equals 8 m/s. In which direction
19:53will it seem to come? In which
19:54direction will it seem to you that the
19:56wind is coming? Is it coming this way
19:57or that way? It will definitely seem
19:59that the wind is coming this way. So
20:01now you will draw it. You drew the wind
20:03relative to the pedestrian. How much is
20:05that? 8 meters per second. In which
20:06direction? It will seem to be coming
20:08this way. This much is easy. This much
20:10is understandable. You can absolutely
20:11feel it. Wind relative to the rider is
20:13higher. In which direction? This way.
20:15That part is done. So, look, we have
20:17calculated the wind relative to the
20:18rider. So what do we have now? Look,
20:20who is there relative to the rider? Who
20:23is it? There is wind. Okay? Who is
20:25there relative to the rider? Wind. Now,
20:28look, with this wind, whose task must
20:30we perform? The rain. So now, we will
20:33work with the rain relative to the wind
20:35. How was the rain falling? It was
20:37falling vertically at 5 meters per
20:39second. Okay? Now think about this, the
20:42rider is here. How should this rider
20:44hold the umbrella to stay protected
20:46from the rain? There is rain and there
20:49is wind. Relative to whom? Relative to
20:51the rider. But who? Who is there? There
20:54is wind. What will the wind do to the
20:56vertically falling raindrops? Won't it
20:57tilt them this way? It will push and
20:59tilt them like this, right? So what do
21:01we have now? Rain and wind. What
21:03happens with rain and wind? I just said
21:04it a moment ago. It becomes the
21:06resultant of rain and wind. So the rain
21:08, pushed by this wind, will tilt and
21:10fall along this path with the resultant
21:12velocity. If the rider has an umbrella,
21:15they will have to hold it at this theta
21:17angle with the vertical. So our target
21:19is to find the value of the theta angle
21:21. So, it's a very simple calculation.
21:23If we calculate the value of the theta
21:25angle, then theta will be tan inverse 8
21:28by 5; if you calculate it, you will see
21:30it comes to 57.99. Degrees. So, look,
21:33the rider now has to hold the umbrella
21:35at a wider angle. The angle before was
21:3821.80, now it is 57.99 degrees. Now
21:42think about it practically; the poor
21:44guy was standing still, rain was
21:45falling vertically, the wind gave a
21:47push, and he tilted the umbrella a
21:49little. But in the direction the wind
21:53is coming from, if he starts running
21:54along the gust of wind, he will tilt
21:56the umbrella at an even wider angle. So
21:59as he tilts it at a wider angle, what
22:01does that angle value become, 57.99?
22:04Degrees. If you want, this math—what
22:06I did first—is the wind relative to
22:08the rider. Meaning, I calculated the
22:10rider and wind first, then did the math
22:12for the rain with that wind. You could
22:15have done the wind and rain first.
22:17After doing the wind and rain, you
22:19could have done the rider's part with
22:20that, but it would have been difficult.
22:22You can try doing that if you want. Why
22:24was this easy? Because the two were on
22:27the same straight line, we calculated
22:29those two first. So let me say it again
22:31, if there is a calculation for three
22:33things—a rider, wind, and rain—then
22:35do the rider relative to the wind
22:37calculation first. Then it will be the
22:38relative velocity of the wind with
22:40respect to the rider, and since that
22:41has become the wind, you will then find
22:43the resultant of that wind with the
22:44rain. Okay! So, let's look at another
22:46math problem of this type. Let's look
22:48at the third math problem. For this
22:49third one, we will do the same previous
22:51problem, just reversing the
22:52pedestrian's velocity. Meaning,
22:53previously the pedestrian started
22:55walking from West to East; this time,
22:57it will be the exact opposite; they
22:59will start going from East to West. So,
23:02let's change it here, from West to,
23:04let's make it from East to which
23:05direction? To the West. This will be
23:12the third math problem. So, you should
23:14note down the third math problem now.
23:16So, let's start solving it. Let's draw
23:20the diagram again; rain is falling
23:21vertically with a velocity of five
23:23meters per second, and the wind is from
23:30East to West with a velocity of two
23:32meters per second. Now, in this case,
23:37what is the pedestrian doing? The
23:39pedestrian is starting to walk from
23:41East to West. With what velocity? With
23:43a velocity of 6 meters per second. So,
23:45let's draw this diagram. This is the
23:46pedestrian. This time, the pedestrian
23:48starts walking this way. With what
23:50velocity? With a velocity of 6 meters
23:52per second. The question is, in which
23:53direction should the pedestrian hold
23:55the umbrella to stay protected from the
23:56rain? So, see, if we could solve the
23:58previous problem, we can solve this one
24:00too. How many are there here? Three:
24:02pedestrian's velocity, rain's velocity,
24:04wind's velocity. Which two are on the
24:06same straight line? The wind velocity
24:08and the pedestrian's velocity are both
24:10on the same straight line. So, I have
24:12to deal with these two first. So, what
24:14happens to the wind with respect to the
24:16pedestrian? Relative velocity. So, the
24:18one with respect to whom the velocity
24:19is measured, their velocity must be
24:21reversed. We will reverse the
24:22pedestrian's velocity. The pedestrian's
24:23velocity will reverse and go this way.
24:25Then we have to find the wind's
24:26velocity with respect to the pedestrian
24:28. After that, we have to do the
24:29remaining work with the rain. So, let's
24:31write it down. So, what is the wind
24:33with respect to the pedestrian? Let's
24:37write here, with respect to the
24:38pedestrian, the velocity of the wind.
24:44So, understand this carefully. This
24:46pedestrian is going this way with a
24:48velocity of 6 meters per second. Just
24:50think about it. Suppose this is you.
24:52You are going this way with a velocity
24:54of 6 meters per second. And this is
24:56your friend. Your friend's name is Wind
24:58. They are going this way with a
24:59velocity of 2 meters per second. So,
25:02how will your friend seem to you? Are
25:04they moving forward or falling behind?
25:06What will it seem like? You are going
25:07at six, which is faster. Your friend is
25:09going at a slower velocity of 2 meters
25:11per second. So, won't it seem to you
25:12that your friend is falling behind?
25:14Along which direction? They are falling
25:16behind in this direction. Can you tell
25:17me with what velocity they are falling
25:19behind? Yes. Six minus two, which means
25:21they are falling behind at 4 meters per
25:24second. See, you have found the
25:27relative velocity of the wind with
25:28respect to the pedestrian, whom I
25:30called your friend. Is that correct?
25:33Exactly the same thing needs to be done
25:34here. So if you want to follow the
25:36standard rule, even if you don't fully
25:37understand it, you can still do it.
25:39Reverse the velocity of the object you
25:40are calculating relative to. So take
25:42this six, and reverse its direction.
25:44Now you've reversed it to this side.
25:45This is six. And in which direction is
25:47the wind speed? Along this direction.
25:49How much is it? Two. So, if you find
25:51the resultant of six and two, which way
25:53will it point? It will point this way.
25:55How much will it be? It will be four
25:56meters per second. How did you get four
25:58meters per second? You got it by
26:00subtracting two from six. So, we found
26:03the magnitude of the relative velocity
26:05and its direction. For the cyclist, it
26:07will seem as if the wind is blowing in
26:09this direction. Clear? We understood
26:12this part. So now, who is there
26:14relative to the cyclist? Who is it? The
26:16wind. And what else is there? The rain.
26:18What happens with the wind and the rain
26:20? They form a resultant. Meaning, how
26:22was the rain falling? Vertically, like
26:24this. So now we have the rain and the
26:26wind. The wind will push the vertically
26:28falling raindrops and tilt them, right?
26:32So, due to this tilt, the resultant
26:34rain will fall along this line. The
26:36cyclist was moving in this direction.
26:38This is the cyclist. If the cyclist has
26:41an umbrella, how will they hold it now?
26:44They will tilt it at an angle to the
26:46vertical. They will tilt it at this
26:49angle. So look, they were moving this
26:51way and tilted the umbrella in this
26:53direction. We need to find the value of
26:54this angle, theta. Let's see if we can
26:56calculate the value of theta; I hope we
26:58can. Tan inverse, the opposite side to
27:00theta is the perpendicular, which is
27:02four. Then how much will this be? It
27:03will be five. So, 4 over 5. If you
27:06calculate it on a calculator, it will
27:08come to 38.65. Degrees. So now the
27:10cyclist must hold the umbrella at this
27:12angle. So you see, even with the
27:16cyclist's speed, wind speed, and rain
27:18speed all together, we can find out
27:19which way to hold the umbrella. There
27:22is another fun math problem here,
27:24number four. The fun problem is number
27:27four, write it down. If the cyclist at
27:33a speed of 2 meters per second from the
27:38east moves to the west. Then how will
27:45they hold the umbrella, let's see if we
27:51can figure it out. So, we will draw the
27:53diagram again; do not try to do the
27:55math without a diagram, okay? So this
27:57is the rain falling at five meters per
27:59second; sorry, the rain is falling. And
28:04the wind is blowing from east to west
28:06at 2 meters per second. Now what is the
28:10cyclist doing? The cyclist is walking
28:12through this rain. Which way are they
28:14walking? They are walking in the same
28:16direction the wind is blowing; that is,
28:18look, from east to west, at what speed?
28:21They are walking at 2 meters per second
28:23. Now look, in this case, which way
28:24should they hold the umbrella? Look,
28:26isn't this problem just like the
28:28previous one? Now look, let's start
28:29solving it. There is rain, there is
28:31wind, and there is a passenger; there
28:33are velocities for all three. So, let's
28:34first calculate the velocity of the
28:36wind relative to the passenger, because
28:38the passenger's velocity and the wind's
28:39velocity are along the same straight
28:41line; calculating these two will be
28:42easier for me first. So, let's
28:44calculate the wind relative to the
28:45passenger and see. Wind relative to the
28:49passenger. Tell me, what will the
28:51velocity of the wind feel like relative
28:53to the passenger this time? Look, let
28:55me give you the same example. This is
28:56you moving this way at what velocity?
28:58At a velocity of 2 meters per second.
28:59And your friend, the wind, is moving
29:01this way. In the same direction. At
29:02what velocity? At a velocity of 2
29:04meters per second. So, to you, the
29:06velocity of the wind. Meaning, will
29:07your friend's velocity feel like
29:09anything else? It won't seem like that
29:10to you; it will seem stationary
29:11relative to you. Because your velocity
29:13is 2 meters per second, and its
29:14velocity is also 2 meters per second.
29:16So, you won't feel like it is moving
29:17forward, nor will you feel like it is
29:19falling behind; you will think it is
29:21always in the same place. That is,
29:22there will be no velocity of the wind
29:24relative to you. The wind will seem
29:26stationary relative to you; you won't
29:27feel the wind anymore. So, can't we say
29:30that the relative velocity is 0 meters
29:32per second? There is no velocity at all
29:34. What happened before? You saw the
29:36velocity was some meters per second,
29:38some meters per second. This time, it's
29:410 meters per second. You won't even
29:43feel the wind; it will seem like the
29:45wind is gone. Now look, since there is
29:47no wind, how is the rain falling?
29:49Vertically. There is a resultant of the
29:51rain and the wind. At what velocity is
29:53the rain falling vertically? It is
29:54falling at 5 meters per second. Now
29:57look, there is no wind. So, there would
29:59be a resultant of the rain and the wind
30:01. If rain falls vertically, the wind
30:03comes, pushes it, tilts it, and brings
30:05it down at a resultant velocity, but
30:07this time there is no wind! The way the
30:09rain is falling, the rain will fall
30:10that way. How is the rain falling? It
30:13is falling vertically. So, the rain
30:15will fall vertically. Now, if a
30:16passenger goes this way, it will seem
30:18to the passenger that the rain is
30:20falling vertically on their head. So,
30:22despite having velocity, how will they
30:25hold the umbrella? They will hold the
30:27umbrella vertically. So now, shall we
30:29say how to hold the umbrella? You have
30:31to hold the umbrella vertically. So
30:34look, you cannot hold it vertically if
30:36you are standing still. Only if you go
30:39in the direction of the wind with the
30:40same amount of velocity, you won't feel
30:42the wind anymore. So it will seem to
30:44you that there is no wind. The rain is
30:46falling vertically. It will feel like
30:48it is falling vertically. You will hold
30:49the umbrella vertically over your head.
30:51So, look at the math. Let me say it one
30:53more time. Just feel it. You don't need
30:55to understand all these calculations.
30:56Just feel the matter this way. The rain
30:58is initially falling vertically. There
31:00is wind, but the passenger has no
31:02velocity. So if I am a pedestrian and
31:04standing still, I must tilt my umbrella
31:06against the gust of wind. Now, if I
31:10move against the wind with a certain
31:12velocity, I have to tilt my umbrella at
31:14a steeper angle. Clear? Look, that is
31:18exactly what happened; it needs to be
31:19tilted at a sharper angle. I'm erasing
31:21this, I've erased it. Now let's move on
31:23to the third case. What happened in the
31:24third case? I have turned around and am
31:26moving in this direction at 6 meters
31:28per second. When I move at 6 meters per
31:31second, the wind is blowing from behind
31:33me. So it will feel to me like the wind
31:35speed is two, and my speed is six. So
31:37it will seem to me that the wind is
31:38actually moving away. At what speed is
31:40it moving away? It is moving away at a
31:42speed of 4 meters per second. So I
31:44would have to hold my umbrella like
31:45this. Now observe this, if the wind
31:47blows at 2 meters per second and I also
31:49move at 2 meters per second, then
31:51relative to me, there is no wind. I
31:53won't feel the wind relative to myself
31:54anymore. The rain was falling
31:56vertically, it will still fall
31:57vertically relative to me, and I will
31:59hold the umbrella upright. I hope you
32:01understand. Now, in the four problems
32:04we just solved, if I replace the
32:06pedestrian with a car, it becomes a
32:08different problem. Then, we won't ask
32:11which way to hold an umbrella for a car
32:13, because a car doesn't use an umbrella
32:15. So what will happen? Which window of
32:17the car will get wet and at what angle.
32:19Do you understand? The question will be
32:22which car window gets wet and at what
32:24angle it gets wet. So let's go ahead
32:26and solve this.

About this transcript

This page contains the full transcript of 2.25 ভেক্টর relative velocity math problems | আপেক্ষিক বেগের বৃষ্টি ছাতার সমস্যা- part -03| by Apurbo physics, generated from the public captions YouTube serves with the video. The transcript has 6,288 words across 905 segments, with the original timestamps preserved so you can click any line to jump to that moment in the embedded player.

What you can do with it

Use the transcript to take notes, quote the speaker, build a study guide, generate a summary with ChatGPT or Claude via the YouTube Summary tool, or export it as a timed subtitle file with YouTube to SRT. You can also re-open it in the transcriber to translate the transcript into 100+ languages.

Free YouTube transcript tool

YouTube2Text is a free YouTube transcript generator — no signup, no daily limit. Paste any YouTube link and get the full transcript instantly, with timestamps, click-to-jump, translation to 100+ languages, AI prompts for ChatGPT, Claude, and Gemini, and exports to TXT, SRT, VTT, or Markdown.