1. ## Car Deprciation Figures

Hello,

I really need some help; I can't figure out how to solve this problem.

You purchase an SUV for $26,000.00. A year later the car is worth only$24,800.00. If the value of the car continues to depreciate at that rate,

a. Find the linear equation that determines the value of the car based on the number of years you own it.
b. When will the car be worth $500? And, I have to show my work! Please help! 2. Originally Posted by chickenwing Hello, I really need some help; I can't figure out how to solve this problem. You purchase an SUV for$26,000.00. A year later the car is worth only $24,800.00. If the value of the car continues to depreciate at that rate, a. Find the linear equation that determines the value of the car based on the number of years you own it. b. When will the car be worth$500?

You have the following data points (t,p) where t is the year and p is the price:
(0, 26000)
(1, 24800)

The equation for a line is y = mx + b where y is the dependent variable, x is the independent variable, m is the slope and b is the dependent variable's intercept. In this case, then, we have p = mt + b.

m = (p2 - p1)/(t2 - t1) = (24800 - 26000)/(1 - 0) = -1200. (In units of $per year.) So p = -1200t + b. Now, we know that when t = 0, p = 26000, so 26000 = -1200*0 + b => b = 26000. Thus p = -1200t + 26000 is the price as a function of time. To find when the car is worth$500, we set p = 500 and solve for t:
500 = -1200t + 26000

I get t = 21.25 years.

-Dan

3. Thanks so much!!!

4. Originally Posted by chickenwing
Hello,

I really need some help; I can't figure out how to solve this problem.

You purchase an SUV for $26,000.00. A year later the car is worth only$24,800.00. If the value of the car continues to depreciate at that rate,

a. Find the linear equation that determines the value of the car based on the number of years you own it.
b. When will the car be worth $500? And, I have to show my work! Please help! I suspect that topsquark's solution is the one that you are expected to find. However there is a different approach that can be used which I will show even though it is not what you are expected to use. Depreciation is often modelled as a fixed percentage of the current value per unit time. Then you would have a model of the value against time of the form: v(t)=v(0).e^{-kt}, taking logs (natural logs "ln") of this will give us a linear model in ln(v) ln(v(t))=ln(v(0))-kt Now ln(26000)~=10.1659, and ln(24800)~=10.1186, so k~=0.0473. Hence: ln(v(t))=10.1659-0.0473 t. When the car is worth$500, we have ln(500)=6.2146, so:

6.2146=10.1659-0.0473 t,

or:

t=83.5 years.

RonL

5. Thanks for your help as well; I appreciate you taking the time to show a different way to solve it!