Question

A body moving with uniform acceleration describes 12 metres in the sixth second and 18 metres in 8th second of its motion find the initial velocity and acceleration

Answer 1

We know , Distance covered by a body in n'th second is given by
S(n) = u + a/2 (2n-1)
So , S(6) = u +a/2 (12-1)
12 = u+11a/2 ➡ 2u+11a=24
S(8) = u +a/2 (16-1)
18=u + 15a/2 ➡ 2u+15a=36
After solving the two eqn , we will get
u=-4.5m/s , a=3m/s²

Answer 2

Formula for finding distance travelled in nth second= u + a/2(2n-1)
u+a/2(2(6)-1)=12
u+a/2(12-1)=12
u+11a/2=12
u+11a/2-12=0

u+a/2(2(8)-1)=18
u+a/2(16-1)=18
u+15a/2=18
u+15a/2-18=0

u+11a/2-12=u+15a/2-18
u-u+11a/2-15a/2=-18+12
-4a/2=-6
-2a=-6
a=3 m/s²
u+11a/2-12=0
u+33/2-12=0
u+9/2=0
u=-9/2

u=-4.5m/s
a=3m/s²

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