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```
If the distance covered by a body in nth second is given (4 +6n)Then find the accelaration and initial speed

```
9 years ago

```							Snth = u + a(n-1/2)
= u -a/2 + an
Snt = 4 + 6n             (given)
comparing both equations
u-a/2 = 4 ,  a = 6m/s2
u = 4+a/2 = 7m/s
approve my ans if u like it
```
9 years ago
```							distance covered in nth second
S=4+6n .......1
also
S=u+a(2n-1)/2 .........2
differentiating equation 1 with respect to n
dS/dn=0+6
dS/dn=6 ......3
differentiating equation second with respect to n
dS/dn=0+a
dS/dn=a
putting value of dS/dn
a=6m/s^2
put the value of a in equation 2
and equate equation 1 and 2
4+6n=u+6(2n-1)/2
4+6n=6n-3+u
u=7m/s
```
9 years ago
```							Sn= u+a/2(2n-1)Also, Sn=u+an-a/2    or    u-a/2+an.......1)                                               and, Sn =4+6n..........2)                         comparing 1) and 2) we get :-  u-a/2=4 and, an=6n.                               As, an=6n ;a=6m/s sq.                                            Also, u-a/2=4i.e.  u- 6/2=4 so, u-3=4 and u=4+3 or u= 7m/s.
```
3 years ago
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