This question was previously asked in

APPSC Lecturer (Polytechnic Colleges) Held on March 2020

Option 4 : 13 m/s

CT 1: Growth and Development - 1

46784

10 Questions
10 Marks
10 Mins

__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/s^{2}

__Calculation:__

__Given:__

a = 6 m/s^{2}

The distance covered by the particle in 5^{th} second S_{5} = 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**