The distance traveled by a body in nth second is given by the expression (2+3n) . Find the initial velocity and acceleration . Also , find its velocity at the end of 2 seconds .

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Solution

we know,
Sn = u + (2n-1)a/2
where Sn is distance traveled in nth second.
u is initial velocity
a is acceleration.
given,
Sn = 2 + 3n
= 2 + 3(2n-1+1)/2
=2 +3/2 + (2n-1)3/2
= 7/2 + (2n-1)3/2
compare above equation to this equation
initial velocity = 7/2 m/s
acceleration = 3 m/s^2

now, velocity after 2sec = u + at
= 7/2 + 3*2 = 3.5 + 6 = 9.5 m/s

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