Question

A body covers 10 m in the 2nd second and 25 m in 5th second of its motion. If the motion is uniformly accelerated, how far will it go in the seventh second?

Answer 1

$S_2=10m,S_5=25m$

For distance covered in $n^{th}$ second,

$S_{n^{th}}=S_n-S_{n-1}$

By 2nd equation of motion,

$S_n=un+\frac{1}{2}a{n}^2..................\left(i\right)$

$S_{n-1}=u(n-1)+\frac{1}{2}a(n-1{)}^2......(ii)$

Subtracting (ii) from (1)

$S_n=u+\left(a/2\right)(2n-1)..............\left(iii\right)$

For 2nd second,

$S_2=u+\left(a/2\right)(4-1)$

$⟹10=u+3\left(a/2\right)$

For 5th second,

$S_5=u+\left(a/2\right)(10-1)$

$⟹25=u+9\left(a/2\right)$

Solving above equation we get,

$a=5m/s^2,u=2.5m/s^2$

Putting these in (iii) and n=7, we get distance covered in 7th second.

$S_7=2.5+\left(5/2\right)(14-1)$

$⟹S_7=35m$