The value of a car is falling 10 percent per year so that if Co is?

Question

the purchase price of car in dollars, its value after t years is given by V(t) = Co(.9)^t. At what rate is its value falling when it is driven out of the showroom? how fast is the car depreiciation after year 1?

Answer 1

Half of the answers so far are junk. Professor Maddie's and Jerry's are good.

Answer 2

Your question is meaningless. You are confusing the rate of depreciation and the value of the car after a given time. Better try again.

Answer 3

Vt=Co[.9]^t
V0=Co
Depreciation is zero at the instant
it is driven out of show room
V1=Co*.9
Depreciation after 1 yr is 10%

Answer 4

V(t)=Co(.9)^t

V'(t)=Co * ln(.9) * (.9)^t

When driven out of the showroom, t=0

V'(0)=Co * ln(9) * (.9)^0 = Co * ln(9)

After one year, t=1

V'(1)=Co * ln(9) * (.9)^1 =Co * ln(9) * (.9)

Just plug in the initial cost Co of the car into each.

Answer 5

rate = first derivative in time

V(t)=Co(0.9)^t

V'(t)= Co* ln(0.9)*(0.9)^t

So, plug in t=0 and t=1 for your answers.