Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by the differential equation $\frac{d V}{d t}=-k(T-t)$, where k > 0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is:
Answer & explanation
Correct answer: option 2
We have,
$\frac{d V}{d t} =-k(T-t)$
$\Rightarrow d V =-k(T-t) d t$
On integrating, we get
$\int d V=-k \int(T-t) d t$
$\Rightarrow V(t)=k \frac{(T-t)^2}{2}+C$ ....(i)
Initially i.e. at t = 0, we have V(t) = I. Putting t = 0 and V(t) = I in (i), we get
∴ $I=\frac{k T^2}{2}+C \Rightarrow C=I-\frac{k T^2}{2}$
Putting this value of $C$ in (i), we get
$V(t)=\frac{k(T-t)^2}{2}+I-\frac{k T^2}{2}$
At $t=T$, we get
$V(T)=I-\frac{k T^2}{2}$