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Amit Tapas Grade: 10
        

If the distance covered by a body in nth second is given (4


+6n)Then find the accelaration and initial speed

6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

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

6 years ago
Fawz Naim
37 Points
										

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

6 years ago
Pankaj
11 Points
										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.
										
8 months ago
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