River Boat problems are related to relative velocity as observed by a moving or stationary observer. The velocity of the boat relative to water is equal to the difference in the velocities of the boat relative to the ground and the velocity of the water with respect to the ground. 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. If you are riding a bike in the rain, how would you hold an umbrella? Naveen Kumar. This will restart the app completely and problems may be resolved. The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river. Numerical : Block moving a on Horizontal floor, Quality Numerical 004 : Projectile Motion inside viscous liquid, Quality Numerical 002 : River Boat Problem, Quality Numerical 005 : Avg Speed and Velocity, Quality Numerical 007 : Distance and Displacement, Quality Numerical 008 : Velocity and Acceleration, Quality Numerical 009 : Average Acceleration, Quality Numerical 010 : Acceleration Graph, Quality Numerical 011 : Height of the Tower, Quality Numerical 012 : Coin Drop from Balloon, Unacademy is Indias largest online learning platform. (b) Find the time required to reach the destination. Irodov on the topics of river . To log in and use all the features of Khan Academy, please enable JavaScript in your browser. The second problem focuses on the launching of a lifeboat in the calm water. Would you point it vertically upwards?
IIT JEE - River Boat Problem Practice Concepts Explained on Unacademy Tap Memory Empty cache . Keep them relevant to the topic. This topic is for class 11 physics students and it comes under kinematics topic. If we look at all the units, they actually do turn out with you just ending up having just meters cubed, but let's do the math. Example: In still water, Peter's boat goes 4 times as fast as the current in the river. Open the " Settings " app on the device.
Relative velocity: Boat problem - GeoGebra Khan Academy's 100,000+ free practice questions give instant feedback, don't need to be graded, and don't require a printer. Our mission is to provide a free, world-class education to anyone, anywhere.
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