HELP ME PLEASE!!!...Determine the velocity and the time.?

Question

A metallic particle is subjected to the influence of a magnetic field such that it travels downward through a fluid that extends from plate A to plate B. If the particle is released from rest at the midpoint C, s=100mm, and the acceleration is measured as a= (4s) m/s2 , where s is in meters, determine the velocity of the particle when it reaches plate B, s=200mm, and the time it needs to travel C to B.

Answer 1

a = vdv/ds

Or, 4s = vdv/ds

Or, 4sds = vdv ------------(1)

Let the particle move from position s1 to position s2. Let its velocity at s1 be v1 and velocity at s2 be v2.

Then integrating both sides of (1)

integral of 4sds from s1 to s2 = integral of vdv from v1 to v2

Or, 4*s2^2/2 - 4*s1^2/2 = v2^2/2 - v1^2/2

Or, 2*s2^2 - 2*s1^2 = (v2^2 - v1^2)/2

Or, v2^2 - v1^2 = 4(s2^2 - s1^2)---------(2)

When the particle moves from C to B, then

s1 = 100 mm = 0.1 m

s2 = 200 mm = 0.2 m

v1 = 0 (because the particle is released from rest at C)

v2  = ?

Substituting the above values in (2)

v2^2 - 0 = 4(0.2^2 - 0.1^2)

Or, v2^2 = 4*(0.04 - 0.01)

Or, v2^2 = 4*0.03 = 0.12

Or, v2 = sqrt(0.12) m/s = 0.35 m/s

Velocity of the particle when it reaches B = 0.35 m/s