 We will start with tutorial problems okay some very simple problems we will take first and from simple all problems are simple only but let us look at these two concept questions okay why are these two concept questions very nice because they will really clarify a lot of doubts okay which a few of you may still have. So what is given is this that two cars okay this is one car this is second car okay they travel along the straight road and what is given is that the position of this cars as a function of time is provided to us and this is the position of first car it is a straight line okay car A this how does the position change okay velocity is 0 here because the slope is 0 velocity increase increase increase okay then the velocity slowly start decreasing decreasing decreasing and it goes. So how does the curve look the slope is 0 the slope increase okay then the slope remains constant little bit then the slope starts decreasing and what we are asked here is that that what are the statements are true that at time 2 t2 both the cars have travelled the same distance okay so look here so at time t2 both the cars have the travelled the same distance they clearly have not because distance is essentially the position and both of them have different positions so they have not travelled the same distance. Second option at time t1 both cars have the same speed now what is speed speed is that the slope of this curve now look at this the slope of this curve B is a little bit larger than the slope of curve A since these two slopes are different at time t1 both cars do not have the same speed so this is not right see both cars have the same speed at some time t less than t1 now this is a very interesting problem in calculus. So for both cars to have the same speed okay what should they have that the slope of this curve okay should be the this is a straight line so this slope should be equal to the slope of this curve at some intermediate point between 0 and t1 will that be true clearly you just take this line and slowly start sliding it parallel to it okay start sliding it sliding it sliding it and you will see that there will be some point here where this line will be parallel to this what does that mean that the slope at this time here okay for this curve and this curve is the same and as a result this statement that both cars have the same speed at some time t less than t1 is right. So what we are doing is that we are just analyzing this curve that take this line keep sliding it parallel to it keep sliding it sliding it sliding it sliding it till it is tangential you will see that at this point this line we are sliding becomes tangential to this which means that the slope of the top line and the slope at this point here will be the same and so they have the same speed both cars have the same acceleration at some time t less than t1 okay. Now what is the acceleration acceleration means okay that acceleration means that the curvature of this that the slope should also change now note here that this speed remains constant okay why because speed is nothing but a slope of this curve which is a straight line. So the acceleration for this thing is 0 whereas you note here in this case you will see that in between 0 and t1 the slope of the curve is always increasing. So the acceleration is always positive so both cars cannot have the same acceleration because this is 0 and this has some finite acceleration so this statement is wrong and the last statement okay both cars have same acceleration at some time t less than t1 and t2 between this now this is true why because note here that here the slope okay the slope is increasing okay then the slope starts decreasing okay. So increase-decrease so somewhere velocity okay so velocity is increasing here velocity is decreasing here okay so somewhere the velocity the acceleration is positive now because velocity is increasing at other part because look the slope is decreasing so the velocity is decreasing so the acceleration is negative and because of continuity between T1 and T2 velocity increases velocity decreases so somewhere the acceleration has to be 0 okay which is the same as the acceleration of this so both cars have the same acceleration at some time T less than T1 less than T2 okay so that much is the information that we can glean from the curves position velocity to position time curves fine now this is very interesting problem so this is the problem that I had discussed okay so this child okay or this person is standing at the midpoint of this merry-go-round now what we do is that okay this merry-go-round we keep on accelerating it such that keep on rotating it such that it reaches a angular speed of omega it reaches an angular speed of omega and at that instant okay the child does not want to stay at the center so he starts walking out okay from the center with velocity with a velocity v u in the outward direction okay and so what is asked to that that if I look from the top what is the direction of the acceleration that the child will face if I look from the top so let us discuss this using the white board what we have is this so if I look from the top this is how the merry-go-round looks like this is the center where the child is standing this is omega or theta dot now let us say I fix the coordinate system like this this is er hat the child is almost at the center okay and this has to be e theta hat now what is the overall acceleration the overall acceleration is r double dot okay minus r theta dot square er okay plus r theta double dot plus 2 r dot theta dot okay into e theta now note one thing what is happening here is that the child starts to move out with uniform velocity okay with respect to this the speed is uniform okay it will so what does that mean is that that r dot is equal to u which is in this direction because the child starting to move out with a uniform speed with respect to this in the direction along the radial direction so this is r dot which is u but r double dot is 0 okay now what is r here at this point at this point just at the origin r is also 0 so this term becomes completely 0 whereas look at this theta double dot note that this is rotating uniformly at an angular speed of omega so theta dot is equal to omega so theta double dot should be equal to 0 so this is gone so what is the term that left 2 r dot theta dot what is theta dot theta dot is this omega what is r dot r dot is the speed in the radial direction okay or the velocity in the radial direction what is that u uniform so what do we have that the speed or the acceleration that the child will face will be 2 u into omega what is the direction e theta now what is this direction from looking from the top it is this direction so we come to this okay and so answer B is the answer because from the top this will be the direction which we will see that the child will face the acceleration and this is the direction that if you try to move out you will tend to be toppled in this direction and this is one example of the so called Coriolis force okay so let us start with some simple problems you are given that the acceleration depends on velocity like this 3-point o 1 v square we are asked to find out v after half lap okay v 0 the car starts from rest so v 0 is 0 r is the radius of the track is 200 meters so the total distance traveled in the half lap is just pi r which is 628.32 now note that acceleration is a function of velocity so what do we want we to want to have a relation between velocity and distance so what do we do we write dv by dt as v times dv by dx do this integral okay vdv by integral adv x0 is beginning to x v0 is the beginning velocity to final velocity now what we want what we know is that that we want to find out what is the velocity when this x becomes 6.238 so what do we do we do this integral this integral can be evaluated okay using simple method substitute v is equal to this will just look as this integral from standard integral will look as log of 3-point o 1 v square why because substitute v is equal to v square is equal to x so vdv will be equal to dx and this just becomes integral of d by a-bx which is log of a-bx okay standard integral we do this what is x is goes from x0 it is 0 to final x which is 628.32 solve it okay we will get that this-v square is equal to this value now how do you want to find out that what will be the velocity we take exponent on both sides so we will see that 3-point o 1 1 v square is this this by the way all these things are there in the nodes in the in the slides which I am displaying in the class so when they will be uploaded you can have a look at them if they are not already uploaded solve for v and you are done now the second important question is that how do you determine the maximum speed that the car can reach now the maximum speed that car can reach okay just note that this is acceleration 3-point o o 1 v square now if this quantity is positive the speed can only increase okay the moment acceleration is positive the speed can only increase the speed can only decrease when this acceleration becomes negative or it becomes deceleration and the speed will stop increasing when this acceleration is equal to 0 so essentially if you want to find out what is the maximum speed the car can take we just keep on finding out the point till which the acceleration is 0 because beyond that the speed can only decrease so just substitute a this should be equal to 0 and we can find out that this is the speed okay at which the acceleration becomes 0 or in other words this is the maximum possible speed that this car can reach because any velocity okay which is different than this the acceleration will be negative or in other words what you will see is that that if the velocity tries to increase okay beyond this that the acceleration decreases and the velocity actually has to slow down okay so the velocity can only slow down after that the peak velocity you can get only at this point because the velocity can only decrease after this and that is the maximum velocity or the maximum speed. Second problem is a very easy problem again an example of what we had done slider block A moves to the left okay it moves to the left with a constant velocity of 6 meter per second determine the velocity of block B so what do we do we just put our coordinate system in principle this is a 2 degree of freedom coordinate system that this has 1 degree of freedom x this is other degree of freedom y so 2 but what is there that these are connected by in extensible strings and as a result okay what we note is that this is a 1 degree of freedom problem why because what is the constraint that this portion remains always constant the total length of the string is what x a plus 2 y b plus some constant and this is equal to the length of the string now you differentiate that with respect to time what do we get x dot plus 3 y dot y b dot is equal to 0 and we can immediately figure out that x a is equal to 6 meter per second to vb will be equal to minus 2 meter per second or in other words because the coordinate axis positive was downwards minus 2 meter second means v is positive upwards okay there is one question okay that many colleges are asking and that question is about this concept problem that I had that I had taken from Bear and Johnston 10 the concept problem was if you look from the top what we have is we have a merry-go-round the merry-go-round rotates okay from the top in a clockwise direction now the child starts from the center okay and at that instant he or she wants to move out with a velocity of u in the radial direction okay it starts from the center where the r is 0 and what is asked to us is that that what is the instantaneous acceleration or what is acceleration at that given instant okay when the child starts to move out with speed with velocity u in the outward direction now to figure that out what we do is that from the origin okay we have this unit unit vector in the radial direction we have a unit vector in the norm in the perpendicular direction and this we will figure out what are the components of acceleration for this particular problem you see that the components corresponding to the radial vector are r double dot minus r theta dot square now note one thing that r double dot term is 0 why because we are told here that the child is trying to walk out with a constant speed u okay and that constant speed essentially is a speed in a radial direction relative to this moving merry-go-round okay so that one point I want to clarify that if the child wants to walk with a speed u okay it will be in the radial direction relative to this merry-go-round so r double dot in that case is 0 second thing is r theta dot square now when the child is starting from the origin what is r at the origin r is very small or 0 so this term becomes 0 what does that mean that there is no component okay in the ER direction now let us check if there is any component in the in this theta direction or not to do that let us go to this first term r theta double dot now note one thing that the child because it is moving with the merry-go-round it has the same components of acceleration as the merry-go-round plus there is an additional relative motion the child is trying to carry out what is that relative motion the child is trying to go outwards in the radial direction okay the child is trying to go out in the radial direction at a speed u that is the only additional component it has over this omega okay for this for this merry-go-round so theta double dot is 0 why because theta dot is constant it is rotating this entire merry-go-round is rotating at a constant omega so theta dot is equal to omega theta double dot equal to 0 let us come to the last portion it is to r dot theta dot r dot is nothing okay but the speed with which the kid is trying to move with relative to this merry-go-round in the relative direction so r dot will become just u and theta dot is omega so 2 u omega in the direction e theta will be the acceleration that will be faced by the that will be the acceleration of the kid and what is that direction that direction is in the direction of e theta so if you look at the merry-go-round from the top we see that acceleration is in this direction going to the left and so the correct answer for that concept question is b where the direction is in this direction another question that was asked on the chat why we have taken this yb to be equal to vb note that we have not taken yb equal to vb only thing that we have changed is we have said that differentiate this equation with respect to time and the relative area and the corresponding equation will get will be xa dot plus 3 yb dot all the constants become 0 will be equal to 0 but xa dot okay is nothing but the velocity of bar a okay in this direction and vb dot okay so and yb dot is equal to vb and then we immediately find out what is the corresponding velocity or the corresponding speed of this so it will be 2 meter per second upwards why because this comes out to be negative our sign convention is downwards is positive with respect to the fixed axis so minus implies the speed is actually upwards in we just take 5 minutes to solve this a very simple problem for projectile motion so it is a baseball machine okay we can replace it by cricket machine okay the problem does not change what we have is that the height from where this comes out is 1.5 meters the velocity is in the horizontal direction what is asked is that that it wants to travel a distance of 12 meters okay so at what horizontal velocity v0 should this be thrown so that the the batter can hit it at a distance of 1 meter from the bottom very simple problem okay let us have a go at it we are asked to find out that if this height should be 1 meter then what should be the corresponding velocity here so it is straight forward that you can apply equations of motion in the y direction apply equations of motion in the x direction and then one particular approach will be to determine time t for the projectile to fall to 1 meter okay so find out what is the final velocity that you get here there is a problem in the printing here okay so this solution is taken from Bear and Johnston so y final is 1 meter x is equal to 12 meter y0 is equal to 1.5 meter this is y0 okay so what we have taken we have taken the coordinate system to be here at the bottom so 1.5 meters is the initial y displacement 1 meter is the final y displacement x is 12 meter so what do we know that we want to go from here to here so yf is equal to y0 plus 0 t why because the ball is projected in the horizontal direction so there is no velocity in the vertical direction so yf is equal to y0 so this is not 3.5 this is not 5 just note one thing that this will be 1 1.5 and the difference will come out to be minus 0.5 okay minus half gt square so you will get what is the time and in the x direction what you say is the x is equal to 0 plus vx0 t is equal to v0 t okay and then solve this equation because this distance is given to us speed in the horizontal direction velocity in the horizontal direction remains constant so 12 meter is equal to v0 into this time and you immediately get what is v0 so very straight forward problem there is this one question which many of you had attempted so a bike moves on a straight road starting at the speed 0 so it starts from 0 this is velocity this is time starts at 0 and then the bike can either accelerate okay it can accelerate with the value of 2 meter per second square or it can decelerate at a value of 1 meter per second square or after accelerating when it reaches some constant velocity it can take that constant velocity and it can keep decelerating till at 100 seconds the bike has to come to rest again that its velocity should be 0 again and this trajectory is one sample trajectory which can satisfy this criteria that it accelerate at 2 meter per second square constant velocity at whatever acceleration has happened and then ultimately deceleration at 1 meter per second square and goes back at time t is equal to 100 seconds the velocity is equal to 0 now note that in this case you want to find out what is the maximum distance that is covered you have to note that the distance covered is the area so there are various possible ways in which you can do it but the maximum distance will be the maximum area and the maximum area can only happen when this only continuously accelerates and continuously decelerates that will be the maximum area and I believe this calculation is correct and answer comes up to be around 3300. Now going to problem number 7 what do we have is all the dimensions are given to you in miles per hour and feet per second square so what I have told given here is that this is a problem from Miriam and Craig sample problem in chapter 2 of dynamics 14 problem okay so what we are given is that car A is accelerating in the direction of its motion so this is the direction of acceleration at 3 feet per second square okay so feet meter it does not matter really stay consistent with the units car B is coming in a curve what is the radius of that curve 450 is the radius of the curve what is the speed 30 miles per hour so what we are asked to find out is determined the accelerator velocity and acceleration which car B appears to have to an observer in car A so what we want to find out is relative to this moving frame A what is the acceleration this what is the acceleration of this car B and if car A has reached a speed of 45 miles per hour for the positions represented so acceleration is given 3 feet per second square for this year for this we have to find out in this position the velocity or the speed of this car is 45 miles per hour or correspondingly the velocity is 45 miles per hour in this direction okay note the coordinate system that we are putting here why in vertical direction x in the horizontal direction in words this is the car moving in this direction at given some conversion terms at 1 mile per hour is 44 feet per second we do this what do we know that V of B okay total velocity of B is V of A plus relative velocity of B with respect to A now how do we find out what are the velocities of A and B for the positions continue speed okay so speed of car A is 45 miles per hour converted into 66 feet per second speed of B is 44 feet per second now note one thing we can do this problem vectorially this angle is 30 degree just the speed is tangential to this curve the velocity is tangential to this curve and a magnitude of this one is 44 feet per second so we just put that in here okay so this is 44 feet per second this is a vector corresponding to the velocity of car B this horizontal one is 66 feet per second is a corresponding vector for this okay this is 30 okay so we have put these values here okay so the angle this tangent okay makes with this line is 60 degree from whatever values that are given to us okay this is 30 degree so this has to be 60 degrees and what we can do this magnitude is given this magnitude is given we can apply law of triangles and what is this V of B okay is given by V of A plus V of B relative to A and we can simply use the triangle law and can obtain what is V of B, A it will come out to be 58.2 feet per second and at this angle we can simply apply law of cosines and the law of signs to get this particular value this magnitude is known this magnitude is known law of cosines will give you what is the magnitude of V of B, A and once you know the magnitude you can use sign rule that 44 divided by sin theta okay will be equal to V B, A which we obtain from cos rule divided by sin 60 from that we will get this angle. Now the second portion is we want to find out what is the relative acceleration okay of B with respect to A for that we write down the equation that acceleration of B is equal to acceleration of A again plus acceleration of B relative to A. Now what is the acceleration of A? Acceleration of A is given to us acceleration of A is 3 feet per second square in this direction. Now how to find out what is the acceleration of B? B is moving at a constant speed so there is no tangential acceleration but it can have a normal acceleration given by V square divided by rho where rho is the radius of curvature okay so V square by rho so that is the acceleration and what is the direction? The direction will be along this okay it will be exactly along this and it will make an angle of 30 degrees with respect to the horizontal. Put that in this is the acceleration of B, acceleration of A is 3 feet per second square and what we have seen is that that acceleration of B is nothing but acceleration of A plus acceleration of B relative to A and now this problem we can do in 2 different ways one simple way is that this angle is unknown to us we can use components and we can say that x component okay x component of acceleration of B relative to A is what? Is the x component of this minus x component of this which is 4.4 cos 30 minus 3 feet per second square and you will get what is A of what is the x component of acceleration of B relative to A? What is the acceleration of B relative to Y is simply the sine component of this just for this triangle 4.4 sine 30 so we get both of these components and if you want to find out what is this angle beta what this makes with respect to the horizontal the relative velocity acceleration of B with respect to A we just use sine rule that 4.4 divided by sine beta will be equal to 2.34 divided by sine 30 and from which we can also get the angle so there are various ways in which we can do the problem so the simplest is we start from a fixed frame of reference obtain the absolute accelerations absolute velocities for both A and B and once we do that the relative velocities can just come by writing a relation like this you can do that problem either vectorially or can using components in the x and y direction both are equivalent and whatever suits you okay is the best way in which you can do this problem okay so with this much okay so let us stop now kinematics of particles.