Q. Equation of motion of a body is $ \displaystyle \frac{dv}{dt} = -4v + 8 $ . Where *v* is the velocity in m/s and t is the time in second. Initial velocity of the particle was zero. Then

(a) the initial rate of change of acceleration of the particle is 8 m/s³

(b) the terminal speed is 2 m/s

(c) Both (a) and (b) are correct

(d) Both (a) and (b) are wrong

Ans: (b)

Sol: $\displaystyle \frac{dv}{dt} = -4v + 8 $

$ \displaystyle \int_{0}^{v}\frac{dv}{-4v+8} = \int_{0}^{t} dt$

$ \displaystyle \frac{1}{-4}[ln(8 – 4v)]_{0}^{v} = t $

$ \displaystyle ln(8-4v) – ln8 = -4t $

$ \displaystyle ln\frac{(8-4v)}{8} = -4t $

$\displaystyle ln (1- \frac{v}{2}) = -4t $

$ \displaystyle (1- \frac{v}{2}) = e^{-4t} $

$ \displaystyle v = 2(1 – e^{-4t}) $

Hence the terminal speed is 2 m/s