Question

a point moves in a straight line under the retardation - KV where K is a constant .If the initial velocity is u the distance covered in t seconds is

Answer 1

retardation , a = -Kv , here K is a constant.
we know, retardation = - change in velocity/change in time .

$\bf{\frac{dv}{dt}=-Kv}\\\\\implies\int\limits^v_u\frac{dv}{v}=-K\int\limits^t_0{dt}\\\\\implies[lnv]^v_u=-Kt\\\\\implies[lnv-lnu]=-Kt\\\\\implies ln\frac{v}{u}=-Kt\\\\v=ue^{-Kt}$

Answer 2

Point moves in a straight line under the retardation -

KV where K is a constant stating that d2x/dt2 =-a(dx/dt)^2, dx/dt= ±(1/a)d2x/dt2,

distance covered has meaning for positive values only,thus dx/dt=-d2x/dt2 d2x/dt2 +dx/dt=0,

characteristic equation is m^2+m=0,m(m+1)=0,x=c1+c2e-t,

two initial values are required, initial position=c1+c2=0 (for the sake of simplicity and without loss of generality),

dx/dt=u=-c2, at t=0