Question

a body of mass 3kg moving with a constant acceleration covers a distance of 10m in the third second and lbm in the 4th second respectively.
.What is the initial velocity of the body ?

Answer 1

distance covered in nth second= u+a/2 (2n-1)
u= initial velocity
n= nth second
a= acceleration

distance covered in 3rd second= 10 m
therefore,
10= u+ a/2 (2*3-1)
10= (2u + 5a)/2
20= 2u+5a----eq.1

Similarly,
16= u+ a/2 (2*4-1)
16= (2u+7a)/2
32=2u + 7a----- eq.2

two equations, two variables
Solve it,
you'll get acceleration= 6m/s^2
and initial velocity= -5m/s

Answer 2

Heya !
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in 3th second 

S(3th) = 10 m 

using 
S(nth) = u + a/2(2n-1) 

S(3rd) = u + a/2 (2x 3 -1) 

10 = u + a/2 (5)

10 = u + 5a/2 

MBS by 2 

20 = 2u + 5a ------(1)
Similarly 

S(4th) = u + a/2(2n-1)

16 = u + a/2(2 x 4 - 1)

16 = u + 7a/2

MBS by 2 

32 = 2u + 7a    ------(2) 

By  subtracting (1) from (2) we get ,

(32-20) = (2u-2u) + (7a - 5a) 

12 = 0 + 2a 

12 = 2a 

a   = 6 m/s²

Put value of a in 1st equation we get , 

20 = 2u + 5(6) 

20 = 2u + 30 

20 - 30 = 2u 

-10 = 2u 

-5   = u 

hence u = -5 m/s 

where negative (-ve) sign represents the direction of velocity