Question

Let $I$ be the purchase value of an equipment and $v\left(t\right)$ is the value after it has been used for $t$ years. the value $v\left(t\right)$ depreciates at a rate $\frac{dv\left(t\right)}{dt}=-k(T-t)$ where $k>0$ is a constant and $T$ is the total life in years then the scrap value $v\left(T\right)$ of the equipment is

• A. $I-\frac{k{T}^2}{2}$
• B. $I-\frac{k(T-t{)}^2}{2}$
• C. $e^{-kT}$
• D. $T^2-\frac{I}{T}$

Answer 1

The correct option is A $I-\frac{k{T}^2}{2}$

Given,

$\frac{dv\left(t\right)}{dt}=k(T-t)$

$\int dv\left(t\right)=\int (-kT)dt+\int ktdt$

$\Rightarrow v\left(t\right)=-kTt+\frac{k{t}^2}{2}+c$

At $t=0,c=1$

$\Rightarrow v\left(t\right)=-kTt+\frac{k{t}^2}{2}+1$

At $t=T$

$\Rightarrow v\left(T\right)=-k{T}^2+\frac{k{T}^2}{2}+1=1-\frac{1}{2}k{T}^2$