Calculus (in Physics , doubt).Plz help.?

Question

1. A point moves in a straight line under the retardation k*v*v.

if the initial velocity is u , prove tht the distance covered in t seconds is 1/k*log(1+kut).

Answer 1

I suggest that you write down

a = dv/dt  =  k*v*v

Solve for v,

then write v = dx/dt and solve for x

Answer 2

The accleration is f*v*v so we may write:

a = -kv^2 = dv/dt

kdt = (-1/v^2) dv

Integrate:

kt = (1/v) + C

Now solve for C using the fact that at t = 0 we have v = u

So C = -1/u

Solve for v:

kt = 1/v - 1/u

1/v = kt + 1/u

v = 1/[1/u + kt] = u/[ukt + 1]

If s is the distance function then:

ds/dt = v

And if we integrate this we can get the relationship between s and t.

let z = ukt + 1 so dz = kudt and dt = [1/(ku)]dz

ds = vdt

ds = (u/z) [1/(ku)]dz = (1/k)(1/z)dz

s =  [1/k]ln(z) + D

s =  [1/k]ln[ukt + 1]  + D

s = 0 at t = 0 so

D = (1/k)ln(1) = 0

And finally:

s =  [1/k]ln[ukt + 1]