Scroll down and tap Google Play Store. Naveen Kumar. Rain-Man & River-Boat Problems for Super Achievers Batch. A boat's speed with respect to the water is the same as its speed in still water. Free online courses for neet 2020 Let's learn Rain man and River Boat Problems live menti quiz for (NEET Physics) NEET 2020 exam by Gaurav Gupta, and boost y. Find the rate of the current. This video discusses the theory behind the famous river boat problem. From equation (3.7) , we know t = d/v_y = d / v_ABcos = d/ v_Asin. In this example, we find the resultant velocity vector of a boat. Now for time to be minimum , denominator must be maximum this implies =0 and =90, but this doesn't make sense as when we try to row the boat at 0 degree with y-axis due to stream it will have a some net velocity which will lie in first quadrant . Visit https://www.fundamenthol.com/ for solving more than 250 mock tests, attempt All India test Series and solve adaptive practice questions for both JEE and NEET. Then tap on Clear data . Created by Mahesh Shenoy. (a) Find the path which he should take to reach the point directly opposite to his starting point in the shortest time. Download our apps to start learning, Call us and we will answer all your questions about learning on Unacademy. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. (b) Find the time required to reach the destination. Homework Equations. River boat problem can be solved easily without applying difficult concepts of relativity and vectors. Sort by: Solution: Let x = rate of the current. V equals 8.2 times 10 to the minus 4 cubic meters. Watch Now Share The class will be conducted in Hindi and the notes will be provided in English It will immensely help those preparing for JEE and other exams. The mathematics is easy! Vector Components Vector Resolution Component Addition Relative Velocity and River Boat Problems Independence of Perpendicular Components of Motion A study of motion will involve the introduction of a variety of quantities that are used to describe the physical world. Open the Play Store again and try the download again. The Physics course is delivered in Hindi. This example can be examined under two part vertical and horizontal motion as in the case of projectile motion. Irodov on the topics of river . NEET UG. If he spent 3 hours traveling up and back, which could be used to determine his speed on the way up the river? The mathematics of the above problem is no more difficult than dividing or multiplying two numerical quantities by each other. Dec 5, 2021 1h 51m . Time= distance_y / v_y. River Boat Problem Practice. When an object, say, a boat, travels at a certain velocity, and the medium through which it travels, say, a river, has its own velocity, we can find the resultant velocity of the object by adding the two velocities. The session will be conducted in Hindi and the notes will be provided in English. You want to cross the river from point A (South) to a point B (North) directly opposite with a motor boat that can manage a speed of 5 metres per second. Tap Memory Empty cache . The session will be helpful for aspirants preparing for IIT JEE moving . 2M watch mins. Khan Academy is your one-stop-shop for practice from arithmetic to calculus. Understand the concept of Rain-Man & River-Boat Problems for Super Achievers Batch with NEET UG course curated by Prateek Jain on Unacademy. Open the " Settings " app on the device. Mar 11, 2022 51m . Would you point it vertically upwards? Watch Now. Apr 9, 2015 #8 gracy 2,486 83 Relative velocity of approach /SEPARATION In this. You can ask your theoretical doubts during this Live session. For PDF Notes and best Assignments visit @ http://physicswallahalakhpandey.com/Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, . The second problem focuses on the launching of a lifeboat in the calm water. Khan Academy is a 501(c)(3) nonprofit organization. Solve Motion Word Problems: One object going and returning at different rates. Created by Sal Khan. in this video we're gonna solve a couple of problems on relative motion and then we'll derive a formula a general formula to calculate relative velocities so here's situation number one so we have one person on bike traveling towards the right at five meters per second let's call him Akash then we have a second guy jogging towards the red nine meters per second let's call him bolt and the . Tap Apps & Notifications then click View all apps . He takes a 15-mile trip up the river and returns in 4 hours. Question (reworded): A guy travels up a river at v mph for 2 miles. In this session, Nitin Sinha Sir will cover River Boat. In this session, Naveen Sir will provide in-depth knowledge of River Boat Problems . So, the time is taken by the boat to go 5km in stationary water = 5/4 hrs = 1 hrs = 1 hr 15 minutes Boat & Stream - Sample Questions PDF:- Download PDF Here Candidates must go through the questions given above and start their preparation for the upcoming competitive exams. in this video we're gonna solve a couple of problems on relative motion and then we'll derive a formula a general formula to calculate relative velocities so here's situation number one so we have one person on bike traveling towards the right at five meters per second let's call him Akash then we have a second guy jogging towards the red nine meters per second let's call him bolt and the question now is what's velocity of Akash as seen by bolt with respect to bolt okay let's try to figure that out the first thing to do is jump in two bolts point of view think from bolts point of view well bolt will not see himself moving I mean when you're jogging you don't see yourself moving and therefore from balls point of view bolt is at rest but instead when he looks at the ground he will see the ground traveling backwards at nine meters per second so that's the first thing to remember that from bolts point of view the whole ground is traveling backwards at nine meters per second and now question is what is Akash doing with respect to world to do that we will wait for one second and then we'll see where Akash ends up at the end of one second okay so in one second we will see Akash traveling forward five meters on the ground because he's traveling five meters per second on the ground but in that same one second both Akash and the ground will travel back nine meters this is what's going to happen okay so let's just write that down we see in one second this biker boy goes five meters forward on the ground in one second but in that same time the ground will carry him back nine meters per second and so notice now effectively Akash would end up going four meters backward every second and that's what bolt will see and therefore we can now say velocity of Akash with respect to bolt is four meters per second backwards and the way we you would write this is we will write it as V of a a four Akash velocity of Akash as seen by bolt so with respect to bolt we have to mention that right we're going to write a second letter for that represents with respect to whom we are calculating the velocity of this first fellow that is 4 meters per second backwards now since we have front and back and we have velocity which is it depends on direction let's just use sign convention right it's it's easier to talk in terms of plus and minus rather than forward in backward so let's choose one direction positive well let's choose the right side as positive then velocity of a is positive velocity of B is positive but velocity of a with respect to B V a B this is negative because ball sees this fellow going backwards so this is now minus 4 meters per second all right now the question is can we can we build a formula for this all right so let's try to write this relative velocity V a B in terms of a formula so to generalize this let's say this guy is having a velocity V a towards the right and let's say this fellow is having a velocity V B so the question now is what is V ay B well let's see what we did we what we did over here to calculate relative velocity is we actually did 5-9 okay and that v is VA so 5 minus -9 what is 909 is VB so if you think about it we have now just built a formula look at this carefully the formula is where B equals VA minus VB it's a very easy very simple formula to remember and this formula helps you calculate velocity of a with respect to B we'll come back to this formula a little bit later but first what we'll do now is we'll take another example and let's see whether we can do the same thing with the second example and here is situation number 2 over here we have a snail traveling towards the right and a train throwing traveling towards the left okay and what we are going to do is try to find out what's the velocity of this snail with respect to the train okay what I want you to do is first pause the video and try to figure this out yourself using the same exercise would read over here and when you're doing this please don't look at the science for a while just first forget about the science just do this logically and see if you can come up with the answer all right let's see let's do this logically first the train is traveling towards the left 50 meters per second but once you jump into the Train from the trains perspective while the train is not moving it's at rest instead the whole ground is moving towards the right 50 meters every second and on that ground the snail is traveling 2 meters per second so if we wait for one second what will the snail do where will the snail be well in one second the snail would have traveled 2 meters forward on the ground but then the whole snail and the ground the whole thing would have traveled 50 meters forward okay notice in this case that's happening in the same direction all right so if you put that together let's write it down somewhere let's write it down over here in one second we found the snail travels two meters every second on the ground but in that one second the ground will carry the snail forward 50 meters per second and so notice if you put this all together you would see the snail effectively going forward 52 meters per second and this now is the velocity of the snail with respect to the train so we can now write this with the notation velocity of the snail with respect to the Train is an incredible 52 meters per second the snail is super fast with respect to the train and this is this sort of makes sense because you may have experienced this if you're traveling in a some vehicle in one direction and if you have ever seen vehicles approaching you in the opposite direction you may have seen them zooming very fast with respect to you that's exactly what's going on okay now let's bring in the signs okay if you use science we will see that velocity of the snail is positive let's call this as V s and this is positive velocity of the train well that is negative because the train is traveling towards the left that's negative okay and the relative velocity will that that is positive and so the next thing is to build a formula and again I encourage you to pause this and see if you can do it yourself right so please try to do this let's build a formula let's let's do this let's put them together okay what is this equal to okay what did we do over here or what we did is two plus 52 is the velocity of the snail so velocity of the snail plus plus 50 well what is 50 that's velocity of train oh no no that's not velocity after that is negative of the velocity of the train correct so 50 is negative velocity of train so to this we added negative of VT and so if you now put them together we will CBS T is equal to vs minus VT tada there we have it that's our formula for the second case so we built a formula for relative velocity of of objects traveling in the same direction we now also did that for objects traveling in the opposite direction but but if you look at them carefully you see that we have gotten the same formula V a be equal to VA minus VB vs T equal to vs minus VT ooh you know what this means we can use this as a general formula in any case we want that's very nice this is our general formula this is one last thing to ponder upon is notice in this formula VA and VB are the velocities with respect to the ground but it doesn't have to be ground let's say let's imagine that these guys were not not traveling on the ground but let's say they were traveling on some of some giant treadmill you know it's and some sort of a travel later or something like that you know imagine this they were on this platform and the whole platform was traveling towards the right at 30,000 meters per second now it might seem like the whole situation has become extremely complicated but it hasn't the relative velocity still remains the same the formula still works and the only small difference is V and V B now would be velocities with respect to the platform okay and here's how I like to convince myself that this will work earlier than we did with respect to the ground remember ground is a surface of the earth and earth is actually a giant platform that's going around the Sun so there's nothing special about ground so if it if this formula works for ground reference frame it will also work for any other point of view or any other reference frame like the platform or maybe they are swimming with respect to the river anything will do so to summarize everything this is the general formula to calculate relative velocity between any two objects and when you're using the formula make sure of two things one use proper signs their velocities assign sensitives and second make sure that VA and VB or V s and V T there are velocities with respect to some common reference frame it doesn't have to be ground it can be with respect to a platform it could be respect to a river or air but some common reference frame, Middle school Earth and space science - NGSS, World History Project - Origins to the Present, World History Project - 1750 to the Present, Creative Commons Attribution/Non-Commercial/Share-Alike. 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