Question

A companys car has an original value of $85000 and will be depreciated linearly over 6 years with scrap value of 10,000.
A. Find the expression giving the book value of the car at the end of the year (0(equal to less than) t (equal to less than 6)

Answer 1

is it different because the previous one has a scrap value of 0 and the new one has a scrap value of 10000? On the new one the professor wrote v=c-r(t)
but that confuses me because we didn't do that for the previous one

Answer 2

change in value = scrapValue - orginalValue
change in value = 10,000 - 85,000
change in value = -75,000


change in time = (new year) - (old year)
change in time = 6-0
change in time = 6


we have the item losing $75,000 over 6 years

annual rate of depreciation = (change in value)/(Change in time)
annual rate of depreciation = (-75,000)/(6)
annual rate of depreciation = -12,500


so the car is losing $12,500 a year

Answer 3

book value = (starting value) + (rate of change)*(number of years)

book value = (85,000) + (-12,500)*(t)

book value = 85,000-12,500*t

book value = -12,500*t + 85,000

Answer 4

if I wanted to know the book value at the beginning, then I just plug t = 0 into the book value equation

book value = -12,500*t + 85,000
book value = -12,500*0 + 85,000 ... replace t with 0
book value = 0+85,000
book value = 85,000

as we expect, the book value is $85,000 at the beginning of the car's lifespan

Answer 5

or...if I wanted to know the scrap value, plug in t = 6 since we're going to scrap it at the 6 year mark

book value = -12,500*t + 85,000
book value = -12,500*6 + 85,000 ... replace t with 6
book value = -75,000+85,000
book value = 10,000

as we expect, the scrap value is $10,000 (the lowest we'll go in terms of keeping the vehicle) at the end of the car's lifespan

Answer 6

here's the graph

https://www.desmos.com/calculator/l0wgunxhxn

see the attached image