00:01
Hi guys, now in this film, the acceleration and time graph is given to us.
00:04
What we need to find out in this problem? we need to find out the position of the particle at 4 .8 second.
00:10
This is an acceleration time graph.
00:12
Now if you see this one, what you observe that? this is the maximum of this one is what? this is given to us.
00:19
This one is this value is equal to...
00:28
This is 6 .8 meter per second score.
00:34
I'm so sorry, it is what? 5.
00:36
Not 8 this is 6 .5 meter per second square we need to find that position of the particle at 4 .8 second when x initial is equal to 0 v initial is equal to 0 from there let's find out what is the area of this one area from there is what change in velocity change in velocity that means i consider it v final minus v initial that will be 1 divided by this is 3 this is 1 divided by 2 times 3 and this is 6 .5 this comes out wevert this will be 9 .25 meter per second so the maximum speed here v final is 9 .25 meter per second now from this after 3 second onwards the acceleration is 0 that means that this v final will be constant this will move with a constant velocity for next 1 .8 second so distance travel in 1 .8 second i would say this is d2 is 9 .75 times this is 1 .8 that is what 9 .75 times 1 .8 which means this is 17 .55 meter let's find out distance from 0 to 3 second.
02:43
Here you can say that acceleration as a function of time, that is slope which is about 6 .5 divide by this is 3 times t.
02:59
I can say that dv is equal to 6 .3 times t.
03:08
Let me integrate both side.
03:11
So i can say that this v is equal to 6 .5 times 3 t squared divided by 2 plus constant c now we know that at t is equal to 0 v is equal to 0 that c will be 0 from there i can say that d x is equal to 6...