A body covers 12 m distance in 3rd second of journey and 18 m in 7th second of journey. Find the
distance cover by the body in 10th second of journey.
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We know that distance Travelled by uniformly accelerated body in n
th
s is given by
S
n
=u+
2
1
a(2n−1), where a is the acceleration of the body and u is the initial velocity
it is given that S
n
=4s at n=3
⟹4=u+
2
5a
.............. eq(1)
S
n
=12 at n=5s
⟹12=u+
2
9a
............... eq(2)
solving eq(1) and eq(2), we get
a=4m/s
2
and u=−6m/s
Now we have to find velocity at t=5s
we know that v=u+at
⟹v=−6+(4×5)=14m/s
now distance travelled in next 3s can be calculated by
S=ut+
2
1
at
2
Note: now u=14m/s
⟹S=14×3+
2
1
4(3
2
)=60m
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