Motion in Straight line class 11 physics| v-t graphs | Problems on V-t graphs

0:02 [Music] hey all welcome to homeschool hope everybody are doing fine and welcome back to class 11 physics series i am with the last session on a chapter motion in straight line so we are going to complete the chapter with this video and please do watch my previous video before watching this because this is the continuation of previous video so in my previous video i started discussing about the graphs related to motion right so we mainly have three different graphs uh that we need to study one is x t graph that is a position time graph second one is vt graph that is velocity

0:47 time graph and the third one is acceleration time graph so we have discussed everything about xt graph and in today's video we will talk about vt graph in detail and also some basics of at graph okay see vt graph is the most important graph guys from vt graph you can come to know many many things you know even this is a very very important tool for solving many toughest problems you can come to know so much information about emotion of an object looking at vt graph okay so let's understand what actually we mean by vt graph and initially we will see how vt graphs look

1:36 for different kinds of motion okay see we all very well know that always you know the horizontal axis is your t-axis that is time is plotted in horizontal axis and uh in vertical axis we will take velocity okay fine see if you have a vt graph of this kind okay say the first type of a graph it is so if if a graph okay or or if you have uh if you have a straight line straight line parallel to parallel to t axis

2:24 t axis that is time axis then what does it mean it means that you see imagine the object is here right and this is time one two three seconds four seconds five seconds six seconds etc and this is velocity two meter per second velocity is unit is meter per second right so two meter per second four meter per second six meter per second seven meter per second see what do you understand from this graph you see at time one second velocity was six meter per second at two second velocity is six at three second velocity is six at every time instant the velocity is same right objects velocity is same it's not that object is

3:10 at rest many of them will get confused here you know what in xt graph in my previous video i told if the xt graph is like this it means that object is at rest but this is velocity time graph it means that object's velocity is same right it it doesn't mean that object is rest see object is moving with constant velocity what does this mean okay so what is the meaning of this line parallel to t axis it means that object object moves at constant velocity constant

3:56 velocity okay so that means it's talking about uniform motion it means uniform motion okay so this is really a very very important it means that velocity is constant okay so it doesn't mean that object is rest if you have this particular line in xt graph then it means the object is rest so here don't conclude that object is at rest so this is where you have to be careful guys okay so this is one type of graph and coming to the next type of vt graph you see v in this axis and t in this axis say if i

4:44 have a line something like this okay see here as the time increases our velocity is also increasing right so and it means that this is a graph for uniform uniformly accelerated body accelerated body okay so this is not a constant velocity okay so velocity is not constant here here velocity was constant but here velocity is not constant okay and you know what here v is positive like v is increasing and a

5:31 is also positive okay so just there is a connection with a and vt graph guys you can find out acceleration from vt graph okay so a means acceleration so just as of now remember this much say if if a line is straight line okay so in this direction you know v is positive and a is also positive right so this is a kind of uh a straight line you will get it for uniformly accelerated body okay so acceleration is uniform acceleration is constant but not velocity okay fine and now if you have a graph

6:17 something like this okay so here what do you mean by that it is the body the body shows negative acceleration okay so in this graph a is negative okay so here a was positive and here a is negative and and this is also a graph for uniformly accelerated body only because you are getting a straight line so it means it is moving with uniform acceleration okay if it was non-uniform acceleration you will get a curve okay so if when you get a curve in vt graph okay it means that acceleration a is not uniform

7:03 a is not uniform okay so for non-uniform motion that is if the acceleration is not uniform you will get a curve if the acceleration is uniform you get a straight line always okay so straight line like this means a is positive v is also positive straight line something like this is a is negative that is acceleration is decreasing that is the object is slowing down okay so uh that is what the actual meanings that you must know from the graph so this is a general idea okay so it's very important when you have a line parallel to t axis it is with constant velocity that is this is a graph for uniform motion and a

7:48 non-uniform motion but with constant acceleration this this this or this this is a positive acceleration and here it means a negative acceleration and if you get a curve it means that even acceleration is not uniform okay so this is a very basic idea and now let us understand what is the importance of vt graph pt graphs i mean what information that you can come to know from vt graph just like from xt graph we can find the average velocity and instantaneous velocity right so the slope of x t graph is what the average velocity value so here the slope of vt graph will talk about

8:37 acceleration so we can able to find out acceleration from vt graph and also we can find displacement from vt graph so two important things we can calculate from vt graph one is acceleration the other one is displacement so let us see how to identify that so let me write it first so two important things that we can find out from vt graph that is acceleration we can find acceleration okay and we can also find displacement of an object from vt graph we can find the displacement of an object so first let us talk about how we can find

9:22 acceleration okay say for example let me take this as a vt graph okay so time on t axis and v on this vertical axis see uh for example if this is your straight line you got okay so uh positive velocity and you see this is what your initial velocity okay so this is initial velocity u and this is your final velocity i can say v right fine and you know what is the slope of this line slope of this line is this much so this is the slope of this particular line right see this

10:07 is tan theta how do you calculate tan theta here tan theta is nothing but what is this this difference right so this difference divided by this difference this value right so let me take this is 0 this is 1 2 3 4 5 okay and this let it be a 2 4 5 6 7 okay fine so how do you calculate the slope here slope is nothing but tan theta right so tan theta slope is one and the same so this what is this this is nothing but

10:54 v minus u right so this value final velocity minus initial velocity right so v minus u divided by time right final time final time let it be t2 and this one let it be t1 okay so t2 minus t1 okay will give you the value of slope right and what is this change in velocity v minus u is what change in velocity divided by t2 minus t1 right so time interval so change in velocity by time is what acceleration right so what is acceleration here

11:42 slope slope value is equal to acceleration if you can find the slope of this line definitely you will get the value for acceleration so here can you tell me what is the value for acceleration here what is v v is 7 minus u is what 2 right so divided by you know 5 minus 0 that is 5 so it is 5 by 5 that is 1 meter per second square right so this is what you are acceleration right so just find the slope and you get a value just like in the xt graph we

12:31 found the velocity right with respect to a slope so velocity is nothing but the slopes value similar to that here the slopes value is nothing but your accelerations value so acceleration is what change in velocity by time so change this is the change this is the change in velocity okay divided by change in time okay this length this difference divided by this difference between the two points will give you the acceleration value okay so this is the way of calculating acceleration right so now let us look at one question

13:16 on how to figure out the acceleration from a graph so guys let's understand the question see the vt graph and find acceleration from acceleration from t is equal to 0 to t is equal to 5 and the second question find instantaneous acceleration also so just acceleration means what it is a average acceleration okay so acceleration between two points is what average acceleration and acceleration at the particular time at an instant of time is called as instantaneous acceleration okay so both can be found in a similar way just put a slope i mean just find the slope and value of slope is nothing but the value

14:02 of acceleration okay fine so what they are asking you see just observe uh the vt graph here so uh you know v in this axis and t on horizontal axis right see ah you know all this is one straight line right so initially the acceleration was decreasing actually it is decelerating right and again from this point it is you know the acceleration increases and from here again acceleration increases okay so everywhere you will see sharp increase in increase and decrease of acceleration right so what they asked you to find out acceleration from t is equal to 0 to t is equal to 5 t is equal

14:48 to 0 to t is equal to 5 okay so that means between this point let me call this point as a this point as b this says c this is d this is e okay so you have to find the acceleration between this point and this point so what you will do say 0 to 5 right so if you join a and e right so this is the slope this is the slope this angle you have to find right so just observe the purple color lines say this is your triangle you know this is your slope this angle you have to find so how do you find this angle this this difference divided by

15:34 this difference right so this is your delta v and this much this much this much is your delta t right so what is delta v this minus 30 minus 0 right so i'm not writing it elaborately say as you know you have studied in previous video also how to find a slope right so this this is what this this is v this is v right and this is u right so v minus u v is how much 30 30 minus 0 30 minus 0 divided by time difference is how much from 0 to 5 it is 5 5 minus 0 that is 5 right so it is 30 by 5 so that

16:24 means 6 meter per second square okay so that is your answer fine coming to the answer for second question guys they asked you to find instantaneous acceleration at sixth second so very sixth second that means here at this point time is six so here you put a line that means at this point let me call this point as i at this point you have to find the acceleration okay it is just similar to average acceleration if i ask if i would have asked you to find average acceleration what you will do between this point and this point you know i'll i'll just put a line here this i'll extend right say

17:12 slope of this line will give you average acceleration right see here the average if something is a straight line if you have to find the average acceleration then it is equal to the instantaneous acceleration at any point at any point okay so between e and f okay if you have to find instantaneous acceleration at any point it will be equal to average acceleration between e and f okay so this is what you have to be very careful say i i made this clear in my previous video also so when something is a straight line say finding acceleration means it's between two points or finding

17:57 velocity means it's between two points right say you get the average velocity or average acceleration acceleration in case of vt and velocity in case of x t okay so if something is straight line you know the instantaneous velocity velocity at any instant in this line between ef will be equal to average acceleration between e and f okay so they asked you to calculate instantaneous acceleration but i am calculating average acceleration between e and f that will be equal to this instantaneous acceleration okay so how do you calculate instantaneous acceleration at at

18:43 sixth second right at sixth second so instantaneous acceleration at sixth second is equal to see this divided by this length this is the change in velocity right say what is the final velocity now final velocity is this v here the initial velocity is v because the starting point i am just considering about e and f so e is initial velocity and f corresponding to the f the line drawn is final velocity so this difference is how much it is 40 minus 30 40 minus 30 divided by what is the time gap what is

19:29 the time gap between you see e and f corresponding to t axis it is 8 minus 5 right so 8 minus 5 so that is 40 minus 30 is 10 8 minus 5 is 3 so you would get your answer 3.3 meter per second square okay so this is how from vt graph you can find average acceleration or instantaneous acceleration so the method of calculating average acceleration is similar to instantaneous see if i would have asked you to calculate acceleration between e and f point this is how you would have calculated similar way but here between e and f at any instant if you

20:17 want to calculate acceleration it is similar to average acceleration only between e and f okay so this is really very very important and when as long as the graph is straight line but if it is a curve okay if your graph is a curve then i'll tell you to find the acceleration between a and b then what you would do you would just connect a with b and this is what we call secant right so slope of secant slope of secant will give you value of average acceleration but here if i tell you to find the instantaneous acceleration at this

21:05 point which is coming between a and b then at this point you have to draw a tangent and the slope of the tangent slope of tangent will give you instantaneous acceleration instantaneous acceleration okay but as long as if it is a straight line your average acceleration is equal to the instantaneous acceleration at any point between a and f between this average acceleration between e and f will be equal to instantaneous acceleration at any point between e and f right but but if it is a curve average acceleration is not similar to the instantaneous acceleration between

21:50 any two points okay so average acceleration you will find by finding slope of secant instantaneous acceleration you will find by drawing tangent at the point where you have to find the uh well uh acceleration instantaneous acceleration okay so this is really very important to understand so now we understood how to find acceleration from vt graph it is just by calculating the slope uh between the two points right slope of a line and now let us try to understand how to uh find the displacement from bt graph see guys finding displacement from vt graphics

22:36 very very simple guys it is just area covered under vt graphics displacement if you can find the area under vt graph then that is the value for displacement okay so it's a very very important uh you know tool that you can use for finding uh displacement normally okay so it's nothing but area under vt graph vt graph gains displacement right so let us see how we can find that you see uh imagine we have a bt graph okay v in this axis and the t

23:22 this is a t axis horizontal axis and uh let the time be this is one second two second three second four second and five seconds and this is uh two meter per second velocity in meter per second right so this is 2 this is 4 this is 6 and this is 8 so imagine i have this straight line in vt graph what does this mean that the object is moving with constant velocity not at rest it's very very important this means that object is not at rest it is moving with constant velocity okay so imagine uh this is point a and this is our point b

24:08 okay so if i ask you to find the displacement between a and b okay so my question is find displacement between a and b so how do you find so you will have to just do area covered under these two points find area covered under these two points say if i have to drag project a on x axis that is on t axis okay so here i get and b if i project here so this is what the area covered right between a and b so this is area

24:54 covered covered between a and b so this area which is equal to displacement displacement okay so what is this you know what is this shape it's nothing but a rectangle right so what is area of a rectangle area of rectangle this is a rectangle right so area of rectangle is equal to length into breadth right length into bread so what is the length here t axis this much this much is the length right so 1 2 3 you know 4 minus 1 so this much is

25:39 this this time gap you know on the t axis this is this is what makes your length so it is three into breadth the breadth is this much this much okay you have to check the velocity uh different zero to six that means six meter per second so three zigzag twelve meter is the displacement so area of rectangle itself is displacement okay of an object from a to b so what is your displacement here it is 12 meter okay so the straight line it can be a straight line it can it can be a curve or whatever it is whatever may be the graph you get just area covered under graph if you can

26:26 find then that itself is a value for displacement okay so this is how we will find displacement from vt graphs okay so for example imagine if your vt graph is something like this v on this axis and t on horizontal axis say if this is your vt graph okay so this is object a was having a velocity of this match and uh let me call it as 2 4 right um you know 6 8 okay and object has moved from a to b okay so if i ask you to find the displacement this is a vt graph okay if i asked you to

27:14 find the displacement how would you find so just this area under a b what is the area under a b this much okay so this much is the area under a b so this is area under a b okay fine so this much this this area you are going to find out okay so what is the shape guys it is a trapezium right so your answer is just if you can find area of trapezium then that itself is the value for displacement okay so how do you find what is a formula for finding area of trapezium it is half into sum of two

28:00 parallel sides right sum of two parallel sides right into height into height right so what is uh sum of two parallel sides say uh what what is this side and this side is a parallel right so half into um you know this side this side is how much how much is the length here this side it is 4 okay this is 0 0 to 4 and another parallel side is this so what is this much 0 to 8 right so 8 plus height height is what this much time so if it is 1 2 3 right so 0 to 3 that is 3

28:47 right so how do you calculate this how do you calculate half into half into or you know 8 plus 4 is what 12 divided by 2 plus 3 so that is ones are 6 0 so 6 plus 3 is what it is 9 meters so 9 meters is the displacement uh that you will observe here okay so 9 meters is the area of trapezium you found so 9 meters is the displacement in this case okay and imagine if you have a graph of something like this okay so just if you have a graph like this this okay then how do you find say this is

29:35 v and this is t so it's actually a discontinuous motion motion is not continuous you know what body was started with this velocity okay so at this point it started with this velocity it moved with the constant velocity and then suddenly velocity is changed and it's moving with different velocity but it's constant right so motion is not continuous guys it's not a continuous motion right but then vt graphs are possible something like this so how do you calculate displacement in this case so it's just as simple as this see it is this area right so it is area of this part plus area of this part you have to do area this is a rectangle this is another

30:22 rectangle so area of this rectangle so your displacement s is equal to area of this rectangle plus area of rectangle 2 okay so this is how you will calculate displacement in different situation so if a graph is something like this it's just a rectangle area of rectangle you're finding and if a graph is something like this trapezium shape and you know a trapezium area if you can find you will get the displacement so you may have graph something like this right so it is just our displacement is nothing but area covered under graph is displacement okay so now let's look at few questions so that you will get more clarity see guys on a board i have given

31:10 three questions say graph is given you have to find the displacement okay all three questions is about finding displacement from bt graph so here graph is given here also graph is given but here in a question some information uh you have to find uh the displacement or distance so first you need to draw a graph and then find say let us talk about them one by one say this graph is given you have to find the displacement of an object okay you you see the graph vt graph you can observe here right say uh there is one parallel line right uh it's parallel to t axis and another line parallel to t axis but it's a

31:56 discontinuous motion say this type of motion we will call it as discontinuous it is not continuously moving okay somewhere here and there some stops are there right but anyway so our target is what to find the displacement right so how do you find the displacement area under graph okay so area of this line okay that means this area and this area right so hope you understood what am i talking about this area and this area this is this is one rectangle this is another rectangle okay let me call this as rectangle a rectangle b so here for displacement you have to find area of a

32:44 rectangle plus area of b rectangle will give you displacement okay so what is the area of a rectangle length into bread okay so what is the length here this much this much is your length right so what is your length 1 minus minus of 2 that means you you can understand this is 1 this is 2 this is 3 3 units is the difference so this much is your length that is 3 3 into you know breadth breadth is how much you see 0 to 20 right so uh 20 right so 3 into 20 plus area of b rectangle so what is your

33:30 length length is 1 2 right so 2 units difference so 2 or you can do like this three minus one okay uh into breadth is how much this difference right 0 to 40 40 minus 0 that is 40 right so this is 60 plus 2 for the 80 right so you will get around 140 meters right so what is your displacement s in the sense displacement so your displacement is 140 meters so hope you are getting an idea how to calculate displacement right so if the graph was like this this is how you have to calculate and now you look at this graph here also you have to find the total displacement

34:15 of an object okay you see here this is a constant velocity and here again it is also discontinuous motion okay so this is a line for constant velocity and this is another line for constant velocity okay so just area of this area of this area of this and area of this you have to add this is a rectangle a this is a rectangle b and this is a rectangle c okay so what you have to do area of a plus b plus c you have to do but you see okay say area of c you will get it in a negative sign you know anything below this t-axis

35:02 is negative anything above the t-axis is positive for velocity time graph okay so for velocity time graph so any anything that is above the t-axis is positive anything below the t axis is negative so here when you calculate area you will get it in a negative sign but you may be wondering area can never be negative right area can never be negative correct but that negative sign is talking about the direction okay so negative sign is only for the direction and anyway area of this and area of this will get cancelled right okay see how they will get cancelled you will understand now just observe area of a is how much length

35:48 into breadth right so uh what is the length here 0 to 2 so 2 into breadth is how much 30. right okay so plus b b's area is how much so this is the difference of one unit and this is a difference of 0 to 20 20 right what is this a difference of one unit into what is this 0 to minus 20 right so minus 20 so what do you get here 3 2's are 60 plus 20 plus and you see here you are getting minus 20 okay so this rectangle's area is minus 20 so

36:33 area is minus 20 minus 9 is telling you that this rectangle is there below the t axis it is there in minus axis okay and anyway this and this will get cancelled right so what is your answer answer is 60 meters so that is your displacement here right so when you have something like this looking at this you can understand this get cancelled with this so now what is the area area is only this much so that much you can calculate and that is the answer for displacement okay so you can analyze that just by looking at the graph right fine so this is how you can find out uh you know the displacement from a graph and now look at this question

37:19 read it carefully and try to find the distance traveled by bike okay so they are actually asking distance traveled by by but see you have to plot a graph also here a bike starts from rest accelerates at 3 meter per second for some time and then decelerates at 20 meter per second square till it comes to rest if total time is 20 seconds find distance traveled by by say on using the formula you can find out not a big issue but still you just observe with the help of a graph how am i finding distance travelled by a pipe okay so with the given information i am plotting a vt graph distance i need to

38:07 find out so i am plotting a vt graph okay very good fine see a bike starts from rest okay that means a bike is starting from here so when something is at rest t is equal to zero v is also equal to zero so this is the starting point of a pipe okay so this bike uh from rest accelerates at three meter per second accelerates accelerating means what increasing velocity so you will get a straight line something like this okay so what is its acceleration that means this lines slow they have given it off this lines slope is three meter per second square okay then decelerates at two meter per second

38:54 square decelerates means what negative velocity that means the object is slowing down okay so decelerates it is decelerates till it comes to rest okay decelerates decreases decreases velocity until it comes to rest okay it travels and what is this lines slope acceleration means what this lines slow okay so what is that 2 meter per second square okay so all that will happen so object coming at this point again from this point to this point you know this will happen within a time gap of 20 seconds okay so this is the graph that i plotted and what i need to find out

39:41 find the distance traveled by a bike that means displacement okay if i have to find the displacement here i just have to know i just have to find out the area of this triangle okay so what is the area of triangle so here distance s is equal to area of triangle let me call this as point a point b and point c okay so area of triangle a b c okay so area of triangle abc is how much half into base into height right so this is the formula i have to use okay so here just if you can calculate this you will get distance traveled by a bike guys half

40:28 into base base is what base is this okay base is this distance this you know time gap x axis is what time time gap is how much moving from a to c it is 20 seconds so base you know so here base value you know that is 20 seconds and what you have to find is height we don't know height is what this is your height right this is your height so this is the velocity we don't know height you see 0 to this point if you can know this velocity this this value if you can know you can just substitute in the place of h and you can calculate the area so for finding h so this is your height right okay and you just observe this triangle just observe this triangle

41:16 this triangle carefully slope is given okay so slope is given and consider this as some v consider this as some v then what is this you know we all know we all know right v by something v by something is equal to slope slope is given that is 3 right let me call it as t okay so this is t this is t and we know t plus t is equal to 20 okay right so this is t this is t t plus t is equal to 20 and v by t is equal to slope right so v by t is equal to 3 meter per second so what

42:03 is t t is equal to v by 3 isn't it so what is this is let it be t1 let this be t2 okay so p1 t1 is how much p1 is v by 3 okay plus like that only find t2 okay consider this triangle now this is the slope okay this by this is equal to slope right that is v by v by t2 p2 is equal to slope isn't it so now what is t2 t2 is equal to v by 2 right so t2 is equal to v by 2 so v by 3 plus v by 2 is equal to 20

42:52 can't you find out the value of v here so what is the value of v so what is the lcm i can take here 6 right so 6 in the numerator what do you get 2 v plus 3 v is equal to 20 right so 2 v plus 3 v is 5b is equal to 20 into 6 so 5v is equal to 120 right so v is equal to 120 by 5 so v is equal to how much guys v is equal to 20 uh you know phi phi 2s are 24 right so v is equal to 24 meter per second

43:38 so we got the value of v so that means height v value is nothing but what height right so height we got to know that is 24 meter per second so base of this triangle is time they gave it in question 20 seconds and height we didn't know but using this logic we found the height 24 meter per second right so if you substitute half into base is 20 seconds into height is 24 so ones are tenza right so it is 240 meters so what is the displacement what is the distance traveled by a bike here it is 240 meters is the answer right see

44:24 how plotting a graph has helped us to solve this problem very very quickly usually if you go on a normal logic if you want to find any other formula uh to solve this problem it's a very difficult thing okay you can't find this problem that easily but you see very quickly very easily you can get the answer for this question with the help of a graph so everything is there in understanding a graph so graphs will help you to find the solutions easily and you must definitely know how to plot these graphs okay so that's all about finding displacement from vt graph so now we are going to a last concept that conversion of vt graph into xt graph xt

45:10 graph into vt graph if you know xt graph you can plot vt graph if you know vt graph you can plot xt graph okay so this is a very very important idea one must have about let's learn this guys this is really the most important topic concentrate so you must know few fundamentals here suppose i'll give you xt graph okay so your xt graph looks something like this so what do you mean by this xt graph see position is constant that means object is at rest okay so when the object is at rest what does this mean the object is not having any motion so its velocity is

45:56 zero so for this graph you will not get any vt graph you can't plot any vt graph so if i have to convert this xt graph into vt graph it is not at all possible okay so you can't can't plot vt graph okay you can't convert this into vt graph because object is at rest okay object is at rest fine and now this is our first point you have to remember coming to second point

46:41 in case if let me take it quite a bigger one if our xt graph looks something like this okay so what does this mean this means that uniform motion yes or no we studied xd graphs in our previous video i explained everything in a detailed way so you have to watch that so xt graph when you get something like this what do you mean by that it is a uniform motion uniform motion means you have constant velocity constant velocity okay so this one if i have to convert into vt graph how does a graph looks like so if you have this straight line in xt

47:27 graph then in a vt graph you get the line parallel to t axis yes or no so what do you mean by a line parallel to t axis velocity is constant velocity is not changing actually this line in x t graph is also same meaning no constant velocity right so what i wanted to tell you here is okay if a straight line if a straight line in x t okay then that line how do you represent in vt you will represent that line as okay you will represent you will represent

48:14 that line as parallel to parallel to t axis okay where in vt graph say it's better if you can write all this and keep this one will definitely help you a lot okay fine now coming to third situation guys if in a xt graph i have a curve okay if in a xt graph i have this curve what do you mean by this curve see curve in the sense what velocity is not constant velocity is not constant okay so velocity is not constant velocity

49:01 increases and a is positive a is increasing okay so velocity not constant a is increasing so if you have this kind of a curve in x d graph then you have to show that in vt graph as a straight line okay you have to show it as a straight line okay so you will have a straight line with increasing velocity okay increasing velocity means what here velocity is positive and a is positive velocity is positive a is positive okay so uh remember if you have this kind of a curve it means a is positive

49:49 see in in vt in vt graph increasing acceleration see velocity increases see from here to velocity increases when does an elasticity can increase when you have acceleration increasing when you have acceleration positive okay so this is what the meaning and now coming to situation number four in case in your xt graph okay if you have a curve like this see here when you have a curve like this okay but here when you have a curve like this then this particular curve you have to represent in vt graph as like this decreasing velocity you see

50:36 curve downward curve it always indicates velocity v is negative okay or i can say it in a better way v is decreasing velocity is decreasing okay if v is decreasing a is negative for negative acceleration this is what you will show okay is that clear and what about here what about here see velocity is constant in the sense what here a so in this vt graph if you have this horizontal line parallel to t access means a is always zero okay so but in a vt graph when you have a line increasing

51:21 right so when you have a positive slope for a line a is positive in a vt graph when you have a negative slope say velocity is going this way that means definitely when you calculate a slope you will get a negative value if you get a negative value a is negative after all what is slope here in the velocity time graph slope value is nothing but acceleration value right so this is how you have to carefully think of see guys in the fifth situation in case in the xt graph if i have a curve something like this okay upward parabola right so you know upward bound or whatever it is so how do you represent this in vt graph how do you convert this in vt graph guys

52:09 you see here guys from negative side the object is coming and it is reaching to t axis right so again from negative side only i need to come and you see here if this is a curve if this is upward bowel curve in the sense if it is upward curve you get a positive velocity you get a straight line like this you don't get a straight line like this you will get a straight line like this okay so upward so it's an upward curve right you get a straight line but it should be from negative from negative it should touch zero at t axis see this point is zero right at t axis okay so it should touch at t axis and then go that way from negative side it should start so this is the negative side of vt graph okay so from here i'll start something

52:54 like this it should go positive velocity positive velocity positive acceleration so here a is positive okay with positive acceleration because the curves shape you must see if it is upward bowel okay if the bowel if it is just a shape of a bowel then a straight line must go this way okay velocity must be increasing acceleration must be increasing right so velocity value has to be big big big big okay and if it is inverted bowl if the bowl is something like this if a curve looks something like this then the straight line must go in opposite direction velocity must decrease decrease decrease that's it a must be negative okay so these are the

53:39 fundamental idea one must remember while converting x t graph into v t graph or v t graph into x d graph okay so this is a basic idea with this idea let us see one example how we can convert x t graph into v t graph see guys uh on a board i have an x t graph okay so you should convert this into v t graph so can you all try can you all pause the video and uh try yourself okay you can just pause the video and try yourself with the basic idea that i gave you right so now with five points all the basic ideas i gave you if the x

54:25 xt graph has what how to convert it into vt graph right if you have a parallel line how you have to represent in vt graph if you have a straight line if you have a curve right so keep that idea and pause the video and do this okay so after trying you just observe this you see from okay it is starting at rest in the xt graph the object is starting from rest okay so i have to in my vt graph also i have to start from rest okay from rest to this point you observe it's a upward parabola right it's a curve it's a upward parabola so if it is in a xt if it is a upward parabola then how must

55:12 be yours what must be uh the way you represent in vt you will represent a straight line right positive velocity with increasing velocity since it has started from rest from rest itself you know from point okay let me mark this as point a point to be point c point d and point e right so my object from rest it's a curve upward uh bowel right so increasing velocity it's moving this way so a to b this is what the straight line i get in vt graph okay and now from this point to this point you observe guys this pointed to this point

55:58 see it's a uniform velocity right it's uniform right it's a straight line it's it's a curve curve always convert into straight line if it is a straight line you will represent it as parallel line to the t axis right so because you see from b to c you have a straight solid line that means from b to c here you will have a parallel line so this is from b to c and now c to d you observe guys downward parabola right so this is this is how you have inverted bowel in the sense what the velocity is actually decreasing velocity is decreasing right so that

56:43 decreasing velocity how do you represent and remember decreasing velocity decreasing acceleration means velocity is negative so here if your curve is something like this means v must be negative it must be decreasing decreasing means sharp straight line to the bottom isn't it it must it must cross this t axis because velocity is negative when when is your velocity can be negative when it is below t axis when it can come below t axis right so this is negative y axis so on y axis you have v so this is your negative y axis velocity must become negative right because you see the curve is decreasing from this point to this point it is decreasing that means you know the velocity must be decreasing you

57:29 have to show a straight line uh pointing like this and it should be negative negative means little you have to draw black below t axis below t axis right okay and that's what is from c to d this is from c to d okay now you observe from d to e guys again it's a straight line right again it's a straight line but it is negative right velocity is decreasing so here see velocity decreases negative slope you will get so in xt graph slope slope of a line means velocity so velocity is negative you will get negative slope so in the negative side only since it's a straight line you will show a line

58:15 parallel to t axis right see this this graph this xt graph if i transfer into vt graph you know with the same information this is also talking about the motion of an object and and the same motion is described by vt graph also but in a different way right so this is how you have to convert x t graph into v t graph and v t graph into x d graph okay so i know uh this is not that very easy you have to practice this you must have fundamentals and more questions when you practice you will become perfect in this concept okay but definitely understanding graph is really a very very important thing to solve problems easily okay so that's all about

59:03 the chapter motion in straight line hope i have given enough justice for you i covered all important concepts and i have shown you uh with the questions how to solve it and what concept to apply right so hope this has given enough benefit don't forget to post your opinion in the comment section so see you with the next chapter thank you so much do subscribe our channel to learn the concepts in the easiest way and in a detailed way all the best [Music]