A particle moving with an acceleration of 6 m/s² covers 40 m during 5^th seconds of its journey. What was the initial velocity of particle?

Concept:

The distance covered by an accelerated particle in  second is given by,

$S_n=u+\frac{a}{2}\left(2n-1\right)$

where u = initial velocity of the particle, acceleration of the particle, a = 6 m/s²

Calculation:

Given:

a = 6 m/s²

The distance covered by the particle in 5^th second S₅ = 40 m

$S_n=u+\frac{a}{2}\left(2n-1\right)$

$40=u+\frac{6}{2}\left(10-1\right)$

u = 13 m/s