0:00
Hi everyone.
00:01
We have a question in which that is given that a particle which is experiencing a constant acceleration of for 20 seconds.
00:09
That means in the question they have given that a particle is experiencing a constant acceleration.
00:17
Constant acceleration is experienced by a particle.
00:23
Acceleration for 20 seconds.
00:27
That means for the first 20 second of the motion, the acceleration is constant.
00:30
Constants let's say the acceleration is a okay then this is also given in the question for the for traveling the distance of d1 the first distance d1 in 10 second and another distance d2 in another 10 second so we need to find the relation between d1 and d2 that means a object which is traveling first d1 first d1 distance in uh basically 10 second and then another another d2 distance another d2 distance in another 10 second so we can say that in both the motions acceleration is going to be same that is a because they are coming under the 20 second because for the first 20 second of the motion the acceleration is same so what we need to find we need to find the relation between d1 and d2 so relation between d1 and d2 we need to find in this question let's see how can we do this question? so we will apply the equation of motion, the second equation of motion, and we can just solve for d1 and d2 and then find the relation.
01:47
So as we know from equation of motion, the second equation of motion, ut plus half a t square is the displacement of anybody.
01:54
Now since initially the first body is starting from the rest, so we can say that since u is equal to 0 in the first case, that means for the first case, if i will find for the first case, that means for the first s is d1 and which is equal to u is zero initially plus half acceleration we have taken a and time is 10 square okay then for the second part so from here we can say that d1 is coming out to be 50a this is let's say one now for the second motion what we can do we can just another thing apply so for the second motion for the second another 10 second d2 is equal to now in this case v is going to be equal to v that means the final velocity of the first motion is the initial for the second motion so vt plus half 80 square and now since we don't know the value of v so how can we find the value of v from the first equation of motion which is equal to v is equal to u plus 80 and since u is zero so we can say that v is equal to since for first time is the 10 second that means we consider 10a so v is coming out to be 10 a from the first part so we can say that d2 is equals to 10 a multiplied by another time is also given that 10 plus half a multiplied by 10 square so from here but we can say d2 is equals to it is hundred a plus then 50 a which is totally equal to one 50a...