Question

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

Answer 1

For a body traveling in a straight line with a uniform acceleration:
Distance traveled during nth second = u + (n - 1/2)* a
Given is = 2 + 3 n
Comparing them we get
a = 3 m/s^2.
****We can't compare if distance is a polynomial of degree 2 or more.

u - a/2 = 2. So u = 7/2 m/s
v at t=2 sec, is 7/2 + 3*2 = 9.5 m/s.

Answer 2

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