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Let u, v and a be the initial velocity, final velocity and acceleration of the body respectively.
As we know some formulas that I’ll use here:
a) Sn=u+a/2 * (2n-1) ………(1)
where Sn= distance in nth second and it is only one term here not S*n it should be remind.
b) S= ut+a/2 * t^2 ………….(2)
So, we are going to solve our problem.
as we have to find the distance in 9th second so it is clear here that we will use the first formula (a) here. but the main problem is that we are not given u and a. so now as there are 2 variables that we have to find and this is possible only when we have two simultaneous equation involving only two variable that is a and u.
so now read question again and find the equations to find the a and u.
As in question it is clear that body covers 2 meter distance in 2nd second.
so by using first formula (a)
here n=2 & Sn=2 by putting values we get
2=u+a/2 * {2(2)-1}
2=u+a/2 *3
2=u+3a/2
multiply by 2 on both side
4=2u+3a …………..(3)
and for the second equation now read again the statement that distance covered in next four second is 6 meter.
so we know that it (distance in next four seconds) is the distance that is covered by body in 2 to 6 seconds interval. and we can say this same statement that it is the distance that is the “ Difference of the distances covered in 6 seconds and 2 seconds.”
so in mathematical statement you can write it as:
S(2 to 4)=S6-S2
6= (S in 6 seconds) - (S in 1st 2 seconds)
by using the second formula (b) to find distances in 6 and first 2 seconds.
6=(ut+a/2 * t^2) - (ut+a/2 * t^2)
6=[{u(6)+a/2 * (6)^2} - {u(2)+a/2 * (2)^2}]
6={6u+a/2 * (36)} - {2u+a/2 * (4)}
6=(6u+18a) - (2u+2a)
6=6u+18a-2u-2a
6=4u+16a
dividing by 2 on both side we get
3=2u+8a …………..(4)
so now (3) and (4) are two simultaneous equation by solving, we get.
u=2.3 m/sa= -0.2 m/s^2
so now using our above formula (a) we are able to calculate distance in 9th second.
Sn=u+a/2 * (2n-1)
here S9 means distance in 9th second.
S9=2.3+(-0.2)/2 *{2(9)-1}
S9=2.3-(0.2)/2 *{2(9)-1}
S9=2.3-(0.1)(17)
S9=(2.3–1.7)
S9= 0.6 Answer.
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