### Relative Motion

When we talk about velocity, we first determine a reference point and accordance with that reference point we find the velocity of an object. For example, suppose a person is traveling in a train and it is going with the velocity of 80 m/s to the north, then another train passes it with a velocity of 100 m/s traveling in the same direction. Now the person sitting on the first train will feel that he is going backward towards south or when both the trains will have the same velocity, the person will feel like not moving at all. All these experiences of the person is a result of relative velocity.

So simply we can say that the velocity of an object moving towards a particular direction with respect to another object moving or at rest is called as relative velocity. And the motion at which the object is moving is known as **relative motion**.

The relative velocity formula as per relative motion can be written as: **Vx,y = Vx,g − Vy,g**

Here Vx,y is the velocity of x relative to y; Vx,g is the velocity of x relative to the ground; Vy,g is the velocity of y relative to the ground.

So we learned that relative motion is simply observing an object’s motion with respect to another object at rest or at motion. Like a person who is standing on the top of a hill is at rest with respect to the hill, but he or she is at motion with respect to the sun or the planets in our solar system. To get a clear idea of relative motion and the relative velocity check the example given below.

**Example: **

A river with a width of 600 m is flowing with the velocity of 4 km/hr. Now a man standing at one end of the riverbank wants to cross the river. How much time will the man take who is at a velocity of 5 km/hr to cross the river, given that there is no drift between the start and end point?

**Solution:**

Let us assume that the velocity of the man (Vm) with respect to the river is x î + y ĵ

Therefore, Vm = Vm,r + Vr Vm = (x+4) î + y ĵ

Now as observed from the riverbank the velocity along x – the direction is equal to zero. (Since it was given that there is no drift between the start and end point)

Therefore, x = −4km/h ……… **(i)**

And also, x2 + y2 = 25 ……….** (ii)**

Solving equation (i) and equation (ii) we get, y = 3 km/h Vm = 3 ĵ

Now taking in consideration the motion perpendicular to the course of the river we can say, s = 0.6 km and v = 3 km/h.

So from the above information, we can say that the time will be t = 0.2 hr.

*Therefore the man will take 0.2 hours to cross the river.*

Check the above video to learn in detail about Relative Motion in a more interactive way with attractive features like 3D animations and in-air projections. Stay tuned with BYJU’S and learn interesting topics in math and science in a more effective way.

**Video Transcription **

So now we are going to look at a very, very interesting part or an implication of visitors. We will be into the problem, yeah. It’s called the hunter monkey problem. Now what is this? It’s a very popular problem. You might know about this. Let me set up the problem for you. So yes. A hunter is aiming his rifle right at the monkey. Right. So his rifle is pointing, if you draw a line from the rifle to the monkey’s head, it’s a straight line. So it’s aiming right for him. Now in other words if the gun had a laser on it, the laser would be pointing right at the monkey’s head. Now you are the monkey.

Now you are able to see this. You are able to see the laser on your face somewhere and you are going ‘oh my god.’ You know your death is near. You look down and you see that it’s not too high. You can jump and you can let go and land little safely. Now you are thinking, you see that he is going to, he is almost about to press his trigger. So you know by the time you jump he would have pressed his trigger. So him pressing the trigger and your jump will be simultaneous. So, what do you decide? Do you decide to stay or do you decide to let go and fall? That is the question. You know it’s coming right at you. Should you stay or let go? What will your answer be? Think about it for a moment. Yeah? Now let’s proceed. So now, the answer to this question might depend upon, how far away you are, how high you are, many other things, that the question does not have enough information. That could have been an answer. You cannot say the answer to this question without also knowing, say what distance this is. So, let’s fix that distance. Its 20 meters. Yeah. What velocity? That might also be important. Yeah because the bullet just races away right yeah. A velocity of 20 meter per, we will make it 20 root 2 for obvious reasons per second. Let’s say that is the velocity and let the angle of this launch be 45 degrees. Yeah. So you might have had 3 answers.

You would have said I will stay. I will hold on and stay. You might have said, I will let go obviously or third you might have said it depends. Depends on what? Let’s see which of these answers is correct. So the problem basically breaks down to having something shot at you with a velocity 20 root 2 at an angle 45. You are 20 meters away at a height of, now why didn’t we give the height in the previous question? Because the information was enough to figure out what that height is. Yeah because it’s 45 degrees and you know the gun points right at the monkey’s head. Yeah? So high will the monkey be? That also has to be 20 because his angle is 45 degrees. So that is fixed. Then you know that the monkey is going to end up falling down in this line. Correct. So if the monkey gets hit it will get hit only along this line somewhere. In other words, when the bullet’s x-coordinate will be what? Exactly, 20. Yeah. 20 meters. So when the bullet’s x is 20 only the collision can happen. 100%, if at all it happens. Correct. So let’s find out how long that will take. What’s step number 1? As always, resolve the velocity in to two. In this case it will be 20 and 20. And this will go away. Great. Now looks pretty easy. Velocity is 20 meter per second. You are going to take 20, you have to cover 20 meters. What time will it take? Yes! Yeah, so the time will be 1 second. So the bullet will take 1 second to reach this line somewhere.

We don’t know at what the bullet will be. Yeah? Not too sure. But then, if at all the monkey does dive it will have 1 second to live. Yeah, so it’s a happy monkey. Not really right. It has only 1 second. So let’s see. What height the bullet will be when after time 1 second. Yeah. The idea here is the bullet is not beyond the laws of physics and gravity, right. The bullet, even though unusual it looks like it will just cover a straight line, truth is a bullet also will end up curving a little bit because of gravity. So the bullet’s path is going to be something like this. It’s also going to be a projectile. So we have to find out how far the bullet would have fallen. Because we think the monkey would have fallen more, no? Yeah monkey is much heavier, monkey will fall yeah? Of course your understanding will be neglecting air resistance here. Aren’t we? We are assuming but neglecting air resistance. So now let’s answer this question. What height will the bullet be at? Why? What will you do for this? You have the time. All you will do is write the equation and get the answer. So your initial vertical velocity is 20 minus 5 t2. 20 minus half t2. I am writing this really fast because g is 10 and half of g is 5. So that equals, what is your question?

You have your time with you, its 1 second. So 20 minus 5 is 15 meters. Because your t is 1. So this height, you know is 15 meters. Oh wow. This must be surprising to you because a bullet actually ends up dropping a height of 5 meters right. Yeah. Now what’s there that’s left to do? When will the bullet hit the monkey? When both of them are at the same position. So if the monkey is either above this or below this it escapes. Yes, neglect for now that it will hit the body and all that. Okay, let’s say that the monkey is really small. Only the head will hit. So everything else is neglected so it has to be exactly at this point at 15 meters. Only then it will hit. Let’s see how likely that is. What is the monkey’s motion here? The moment the bullet was launched for the same time t, t equals 1 second, the monkey is also falling down. So the monkey’s fall time t equals 1 second, for the monkey. Its initial velocity is, it’s just letting go at zero, and what should we find. The s right? The acceleration we know. So what will your answer be? You will get half into g is minus 10. Into t2 which is 12. So you get minus 5 meters as your displacement. What does that imply for you? Its 5 meters downwards. So if the monkey starts here from a height of 20 meters and drops a height of 5 meters. Oh my god! The monkey ends up getting hit by the bullet exactly. Is that surprising you? Yeah a tragic end of the monkey right? The monkey had 1 second to live. And then it died. But if the monkey had stayed what would have happened? Yeah the bullet would have gone below and it would have survived. So in this case the answer surprisingly is just to stay because the bullet is aimed right at you but the bullet will fall but by letting go you are also falling you are ending up hitting the bullet.

Now you might ask for this case it’s a coincidence. For most cases letting go might be better. You have arranged the numbers very, very intelligently and somehow made it so that the bullet will correctly hit the monkey. It looks like a bad coincidence to me. I will not believe you. I like what you say. It could be a bad coincidence right. Who knows? 20 I have fixed here, 20 I have fixed here, 45, all this. So what if it’s a special case here. To answer this question let’s take a very different approach. Let’s imagine we look at the same thing from the monkey’s point of view. What will the monkey see? The moment the monkey has let go and the bullet was also fired, both of them have various things that are different, right? But one this is common between the two of them. And what is that? The acceleration g downwards. Both of them are falling at an acceleration g downwards and in free fall I told you that if you walk off a building and if you are jumping along and you have a friend with you and both of you are falling, as far as you look at each other it’s as good as neither of you is falling.

The only way you know is by how maybe by the way the air rushes at you or everything else is going upwards and you are feeling all confused. But neglecting air and if you are looking only at each other what will you see? A normal conversation, both of you seemingly at rest. That’s pretty much what’s going to happen to the monkey here. The monkey’s world in the monkey’s frame of reference in other words, gravity can be switched off. Because in the monkey’s world, it’s at rest so you apply it’s acceleration to itself which means to every other body you will do that. So you do that to the bullet as well. You will do that to the Earth as well. Monkey will see the Earth coming up towards it at g. Yeah but that’s not too relevant to us right now, we forget it. So what does it mean? The monkey sees no acceleration for itself or for the bullet. Because their relative acceleration is zero. The crux here is that their relative acceleration is zero because both of them are accelerating downwards.

Then, what do you observe? What will the monkey see? The monkey will just see this velocity being the 20 root 2. The monkey will see it coming right at it and hit. Yeah. In a gravity less world, the same thing had happened in space, what would you have expected is exactly what will happen now. In space what will happen? If somebody aims at you and shoots if it’s aimed exactly at you as a straight line then it will come and hit you because there is no gravity. And in this world now there is no gravity now right. You have added everything. You have taken away the g. There is nothing else over there. So the bullet will take a straight line path and come and hit the monkey. That’s what the monkey will see. Which means here we haven’t used anything. This velocity here is not even required. What have you shown? No matter what if both of them let go at the same time if the bullet is aimed right at it the bullet will hit. Yes.

Now, let’s see if our calculations match through another method. So this velocity is 20 root 2. You know that this distance is 20, this was 20 so this must have been 20 root 2 meters. Right. So your time taken, right. Doesn’t say it in these two frames. Time is still 1 second. And this cannot change. Right. What point of you you see the problem from cannot change the amount of time the monkey has left to live. In the previous case the monkey left and in our frame of reference the monkey took 1 second then you know in 1 second the monkey was hit. Here also it has to take 1 second. The interval cannot change. The time interval. So it’s interesting. We can make it look even more difficult by asking the question if the same thing had happened in the falling elevator what would have happened. Yeah. That’s pretty much the space. Space or a falling elevator are similar frames right. Because inside the falling elevator, the gravity is switched off so it’s as good as.

Why is gravity switched off? You are falling, bullet is falling, elevator is falling, everybody is falling; as far as you are concerned nobody is falling. Except they will get hit after some time but till then for a while you will feel ‘Hey I am flying!’ inside the elevator. Yes? So inside the elevator as well, the monkey will just see the bullet coming right at it and then hitting it. So that’s a pretty counterintuitive result for most of us, right? Because most people think that we should just let go because the bullet will pass way above. That comes from the intuition that the bullets almost travel in a straight line. Now you know that letting go is actually more dangerous than staying. If you stay you might still get killed because the bullet won’t, you know, diverge too much from its distance. Instead of maybe hitting in the head it will hit you somewhere in the chest, or something like that. But if you do let go you will surely get killed, get hit. Yeah?

So we have shown these two here, so staying, maybe the bullet will hit. Letting go bullet will surely hit. Neglecting air resistance. One small caveat is there here though. What is that? We said, if it is pointed at you it will surely hit you following one small condition. I am pretty sure some of you thought about this. And that is that the bullet’s range must be at least equal to the horizontal displacement, 20 meters. So the monkey is here, right. The bullet’s range must at least be this much because otherwise it doesn’t matter whether the monkey lets go or stays because the bullet doesn’t even reach the horizontal length. To hit, it has to reach somewhere on this horizontal line. If it never happens, it will never hit. So if you are also asked about a condition, when the, or what the minimum velocity of the bullet must be for this to happen, we can calculate that by calculating the range to be equal to 20, in this case.

Depending on the question this number will change. And then backtracking to. You will know the angle, its 45. Yeah? To get maximum range you have to shoot at 45, because that, you will finally end up hitting the monkey at the same flat line, right? That’s when your velocity will be minimum. So you have fixed the angle as 45, fix the distance and you can find the velocity that you need, u that you need. Do observe though that the slower and slower your velocity becomes the longer and longer the monkey has to live. Because if you hit it infinite velocity, the monkey will get hit almost there. Which is pretty much what bullets are. Bullets are very, very fast so we always imagine that to happen. But the slower and slower your bullet the longer and longer the monkey has to live. At your minimum velocity the monkey will land and you will hit at the same time.

But any lower than that minimum velocity, the bullet won’t even reach that final point. So yes, not always will the monkey die. There is some hope in the world if the velocity is enough to cross this line then if the monkey let’s go, its doomed. Now using this idea, the idea of the hunter/monkey problem is to hit at relative accelerations. When two bodies are falling, taking the frame of the falling body can actually simplify the problem.

Let’s solve another problem to understand this. Let’s setup the problem now. So yes. You have seen these, play Angry Birds. Seen one blue angry bird that breaks into three, right? Let’s look at that now and let’s impose one more constraint on it. Let us say that the breakings right, the three little birds end up having velocities lined up like this, right. Now what will that imply? That will imply that their horizontal components are the same. The ux for all is a constant. That’s our constraint. Now we are asking – as the 3 birds are flying, if one of them, the bottom most one looks up and sees the top most one, what kind of motion will it see? Now the answer to this question that I got in my classroom was it will not see the top most one, because the second bird in the middle will block it. So I said I agree.

So if it looks up at the second one, the middle one, what will it see? What kind of motion will it see? I will give you some options. Will it see the motion being (a) a projectile, is that possible? So something like this. Yeah? Let’s move to a different page so we will have all our four options. Yeah. One is a motion like this. One is where it goes up and comes down. So the bottom bird will look at the top bird going up and coming down. Yeah? But nothing else. Third option being, it will see the bottom bird going up with an acceleration, and fourth one being it will see it going up with a constant velocity. So here clearly there is an acceleration. Here there is clearly an acceleration, right? It’s going up and coming down. So these are our four options. What is your answer? Now think about that answer, pause for a moment, and then come back. We will understand how to solve this problem. Now the intuition for this problem is the bottom bird sees the top bird, it will look like some kind of a parabola, yes? So it’s difficult to visualize which is why sometimes mathematics might help us. Let’s look at what we have to do right.

Solving problems in relatives is very, very easy. What is the bottom bird’s velocity in the horizontal, some ux. Yeah? We will call it ux1 because it is the first bird. It will have some vertical velocity, uy1. Correct? Because it ended up having some velocity like this. Now bird two, we will use a different color now. Let’s say had a velocity like this, only it will end up above right? But what is common? The ux. ux2 is also the same number but uy2 will be a different number. Otherwise both ux and uy are same, both the birds will be sticking to each other and flying. Correct? It’s not happening. So uy has gone above, that’s why the bird has gone above. So you have, you will have after some time the bird will be somewhere over here. So you know now, by now I am pretty sure that throughout the motion both of these birds will be along the same vertical line. Which is another way of saying they will have, they will not see each other moving horizontally. So the bottom bird won’t see the top bird moving this way or that way, right or left. Yes? That’s 100% fixed. But let’s not do that in any intuitive way. Let’s just do it mathematically. So that we know we are sure. So we first have to calculate our velocities. What about our relative velocity. Relative velocity – in other words the velocity from 2, velocity of 2 rather in 1’s point of view. What’s that going to be? We will first do it in the x-direction. It’s going to be ux2 minus ux1. Same old 1-d motion, relative motion okay? None of the formulas change. None of the formulae change.

So we have velocity of 2 in the frame of reference of 1 in the x-direction to be this. This you know zero. Great. What about velocity of 2 with respect to 1 in the y-frame. Because that’s what you do right? Who cares whether the body is going like this, flying is happening, none of that. We can break it down into two motions. So we will do that now. You will get uy2 minus uy1 right? That will be equal to some number. Even you know that it will be this arrow. Yeah. Let’s call this ΔvyWe are just using some names to make our representation easier. None of this is that important. Right the names I have used. Δvy, some number, but it’s a constant. This doesn’t change. Do you realize that? Because uy1 is a constant, uy2 is a constant. So Δvy is a constant. It’s the difference between the two. Done. What about the acceleration in the x-direction. A2y in the x-direction. Both of them are zero accelerations so it’s zero minus zero equals zero. Along the x does either of the bird have an acceleration? No. Only g acting downwards. So in our y direction maybe we have a difference. But do we? Both are acceleration g, which is what our hunter/monkey problem setup to right? It was a setup to understand this even better.

So the relative acceleration in both directions is zero. Which means none of these can be correct. This can’t be correct. This can’t be correct, this can’t be correct. When I say none of these a, b or c can’t be correct. The only heralded answer, but let’s acknowledge what’s the beauty over here, because when I think about this problem, when the bottom bird is going and the top bird is going up it’s always in a straight line, but we think when it starts to fall down I will see the bottom top bird falling down also. So people tend to say choose this answer in one sense. Where they will say the balls, it will look like the top bird is going up for a while and then down for a while because they imagine when the bottom bird is coming down the top bird is also coming down, no? So it’s kind of not being able to go completely into the bottom bird’s frame. Which is why math is there to help us. So this is the bottom bird’s world. These four numbers are enough. Yeah? Velocity is being zero horizontally. Vertically there is some velocity.

Acceleration in all directions is zero. What kind of motion is this? Body 2’s motion from body 1’s motion, yeah bird 1’s motion is just some Δvy velocity upwards. No acceleration in any other direction, no velocity in any other direction. So what will it see? It will just see the top bird going up with a constant velocity. Now that might be counterintuitive for many of you. It was for me when I thought about it first. Wow! So the bottom bird sees the top bird, it will just see it going up with a constant velocity even when both of them are falling down. Yeah? Of course after the bottom bird hits then things change because the relative acceleration won’t be g anymore. Yeah? So this gives us a glimpse of how we can simplify problems by you taking another frame of reference.

You will see that again and in an incline plane problem. But before that one of the questions we can ask you now is what if we are in 1-d we threw one ball up, one ball down. Then ask when they will collide. We can just make a 2-d version of the problem. I am saying just because it’s hardly any different. So I can give you a problem that a ball it’s going like this, another ball it’s going like that. Let’s say they will collide. They may or may not collide. Yeah? You might have to find out if they collide, how long it might take. But again over here the insight will be their relative acceleration is equal to zero.

So you will enter the ball’s frame, apply relative acceleration is zero. Then calculate horizontally what must happen, vertically what must happen. Yeah? If they must collide the have to be, they can’t be thrown like this right? Where this velocity is large then clearly they won’t collide. Yeah so you can think of some constraints here as well. Let’s start thinking about it together. One of the constraints can be that, see, this velocity right, horizontal velocity and this, they are pointing this way then you are pretty sure that their paths will cross. But then what must happen then? Their vertical velocities you have to check. Yeah? If, let’s say this now, if we have some vertical velocity like this. Some vertical velocity like this such that they are pointing one of the point here, one other point here. Can we say that they have to collide? Think about this question now. If you have infinite depth. If we have infinite depth, will you agree that they have to collide? Think about that question. Why am I claiming this to be true? If this is true, if I can go to the frame of space, because I realize when the relative acceleration is zero I can think of a space or a falling elevator then this ball has to go here. This ball will go somewhere there. Yeah?

Now if these two velocities are matched only then will the collision happen, so we haven’t confirmed. Correct? Why? I could have velocity like this, this passes the line like that and then this can go like that. Even in space it’s possible. In space it’s possible that these two won’t hit. If I shoot one like this, shoot one more like this they might miss. Which means we cannot say in general that there will be a collision so there have to be some conditions that are satisfied. But to satisfy those conditions what I am pointing out to you is you don’t have to think of them as falling balls you can think of them as straight balls flying in space.

So let’s see, let’s just make up a problem. I shoot a ball like this. Right? At some velocity. I shoot another ball at some velocity. What can you forget for now? That there is any acceleration because relative acceleration is zero. So all you will do is, let me call this u1 let me call this u2. I am doing it in a way so that we get a general idea. Yeah, I could make up some numbers here but not required. So you will have u1 cos θ1 here. u2 cos θ2 here. Assuming these angles are the θ2 and θ2. Yeah? So the vertical will be u1 sin θ1 and here it will be u2 sin θ2. So what we don’t need is the central elector here because we have already broken it down keeping it is only like a useless appendage that we are having.

So we have this here. Now what should you calculate? How long will these two take to collide? That’s a very, very easy thing to calculate. Yeah? Now once you calculate that that will just be this distance divided by u1 cos θ1 plus u2 cos θ2. Now I am being a little, little, little imprecise here. If we had represented these with the approximate, with the appropriate signs then those are actually pointed at the negative u2 cos θ2. So it will come down to this, but yeah I am assuming you know that. What I mean is this will be v2 minus v1 the velocity of this as they approach. But because you have written it in the left direction they add up. They are scalar quantities so speeds add up. So you add this over here, this will be your time. What should you find out now? Now the times are equal. Yeah. So some time they will take to come and meet here. This time will not be different for both of them, then what must be constant? So what must be constant? This must be equal to this right? u1 sin θ1, in other words the vertical velocity of ball 1, this velocity and the vertical velocity of ball 2 must be the same, in this case because being launched from the same height. Yeah? Let’s look at a case like that. So their vertical velocities have to be equal if they are being launched from the same height.

So you can think of other constraints as well right? I can make another problem seemingly difficult by throwing two from different heights. Yeah? Nothing different. Relative acceleration is zero. So your problem will break down to something being thrown like this, something up or down or whatever is happening being thrown like that. All you have to do is calculate this way without any acceleration and calculate if the conditions are being met. If they are then you can calculate the time. The time will be your answer.

To cover this horizontal you have to calculate the time and that will be your answer. Again over here you can go to the adaptive and there are problems which test you on these but they are giving you every single thing you need to be able to solve those problems. Relative acceleration being zero, calculating the horizontal, calculating the vertical, checking if they meet what should happen. You are given some conditions for that, so you are good to go.