1. ## differential equation

a particle moving horizontally in a straight line experiences retardation of kv m/s^2 where v is its velocity. the initial velocity is 150 m/s and is reduced to 50 m/s in a time of 1 second. it travels a distance s m in this time.
(i) find the value of k
(ii)deduce an expression for the velocity at any time t.
(iii) show s=100/ln3 m

I was able to work out part (i) by letting dv/dt =-kv i then integrated this and put in the limits I was given in the question and I got an answer of k=ln3
I am stuck on part (ii) . I have tried letting dv/dt =(ln3)v . I then integrated this and put in the limits of initial velocity 150 and final of v . I got v=ln150-ln3t. this is not the right answer. the right answer is v=150e^-(ln3)t m/s

2. ## Re: differential equation

$\dfrac{dv}{dt} = -kv$

$\dfrac{dv}{v} = -k \, dt$

integrate ...

$\ln|v| = -kt + C$

solve for $v$ ...

$v = e^{-kt+C}$

$v = e^{-kt} \cdot e^C$

let the constant $e^C = A$ ...

$v = A \cdot e^{-kt}$

$v(0) = 150 \implies 150=A \cdot e^0 \implies A = 150$

$v=150 \cdot e^{-kt}$

$v(1) = 50 \implies 50=150 \cdot e^{-k} \implies k = \ln{3}$

$v= 150 \cdot e^{-(\ln{3}) \cdot t}$

thanks