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Let I be the purchase value of an equipm...

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 differential equation `(d V(t)/(dt)=-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 : (1) `T^2-1/k` (2) `I-(k T^2)/2` (3) `I-(k(T-t)^2)/2` (4) `e^(-k T)`

A

`e^(-kT)`

B

`T^(2)-I/k`

C

`I-(kT^(2))/2`

D

`I-(k(T-t)^(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Since total life is T, scrap value is V(T).
We have
`(dV)(t)=-k(T-t)dt`
`rArr int_(I)^(V(t))dV(t) = int_(t=0)^(T)-k(T-t)dt`
`rArr V(T)-I = k[((T-t)^(2))/(2)]_(0)^(T)`
`rArr V(T)-I = -k[t^(2)/2]`
`rArr V(T)=I-(kT^(2))/2`
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