The velocity of a body moving along x-axis is given by v="under root"36+2x where v is in m/s and x is in metre.Find the distance travelled by it in 2nd second

Given: ? V = ?[36 + 2x]
? (V^2) = 36 + 2x
? d(V^2) / dt = d(36 + 2x) / dt

Here, I am differentiating LHS and RHS separately:

LHS
= d(V^2) / dt
= [d(V^2) / dV] * [dV / dx] * [dx / dt]
= 2V * [V (dV / dx)]
= 2V * A

RHS
= d(36 + 2x) / dt
= [d(36) / dt ] + [d(2x) / dt]
= 0 + 2[dx / dt]
= 2V

Thus
? d(V^2) / dt = d(36 + 2x) / dt
? 2V * A = 2V
? A = 1

Now, if we integrate acceleration to get velocity as a finction of displacement, we get:
? (V^2) / 2 = x + C [ where C is constant of integration ]
? (V^2) = 2x + 2C

But, it was given that (V^2) = 2x + 36
? C = 18

Hence, at x=0, V = 6 or V(x = 0) = 6


Let us consider the point where the particle started motion to be the origin.
? at t=0, x=0 or x(t=0) = 0
? at t=0, V=6 or V(t=0) = 6
We had:
A = 1

If we integrate acceleration to get velocity as a function of time, we get
? V = t + C [where C is the integration constant]

As V(t=0) = 6, we get C = 6
? V = 6 + t

Now if we integrate velocity to get displacement as a function of time, we get:
? x = 6t + [(t^2) / 2] + C

As x(t=0) = 0, we get C = 0
? x = 6t + [(t^2) / 2] 

Now for t = 2
x(2) = 14

For t = 1
x(1) = 6.5

Hence, distance travelled in the 2nd second = x(2) - x(1) = 7.5