2) Suppose the value of a car depreciates linearly with time. A new car is worth $41,000; 3 years after being sold, the car is only worth $25,000. Let t = number of years since the car was sold, and let V(t) = value of the car (in $). a) Find a linear function for V(t) that fits the above data. b) What will the value of the car be 5 years after it was sold? c) During which year will the car's value be at $8,000?
2) Suppose the value of a car depreciates linearly with time. A new car is worth $41,000; 3 years after being sold, the car is only worth $25,000. Let t = number of years since the car was sold, and let V(t) = value of the car (in $). a) Find a linear function for V(t) that fits the above data. b) What will the value of the car be 5 years after it was sold? c) During which year will the car's value be at $8,000?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Car Depreciation Problem Set
#### Problem Statement:
1. **Context:**
Suppose the value of a car depreciates linearly with time. A new car is worth $41,000; 3 years after being sold, the car is only worth $25,000. Let \( t \) = number of years since the car was sold, and let \( V(t) \) = value of the car (in $).
2. **Questions:**
a) **Linear Function Determination:**
- Find a **linear function** for \( V(t) \) that fits the above data.
b) **Value of the Car After 5 Years:**
- What will the value of the car be 5 years after it was sold?
c) **Determining Year for a Specific Value:**
- During which year will the car’s value be at $8,000?
#### Detailed Breakdown:
**a) Finding a Linear Function:**
To find a linear function \( V(t) \), we will use the information given:
- Initial value (at \( t = 0 \)): \( V(0) = 41000 \)
- Value after 3 years (at \( t = 3 \)): \( V(3) = 25000 \)
The general form of a linear function is \( V(t) = V_0 - kt \), where:
- \( V_0 \) is the initial value
- \( k \) is the rate of depreciation per year
We can calculate \( k \) as follows:
\[ k = \frac{V_0 - V(3)}{3 - 0} = \frac{41000 - 25000}{3} = \frac{16000}{3} = 5333.33 \]
So the linear function \( V(t) \) becomes:
\[ V(t) = 41000 - 5333.33t \]
**b) Value of the Car After 5 Years:**
To find the value of the car after 5 years, substitute \( t = 5 \) into the linear function:
\[ V(5) = 41000 - 5333.33 \cdot 5 \]
\[ V(5) = 41000 - 26666.65 \]
\[ V(5) = 14333.35 \]
Therefore, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b9391ec-01c2-4ef9-be82-9d31d6150e62%2Fa8df1e56-3758-4fe3-bf45-9c0fb62eaaee%2Ftrxcq6m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Car Depreciation Problem Set
#### Problem Statement:
1. **Context:**
Suppose the value of a car depreciates linearly with time. A new car is worth $41,000; 3 years after being sold, the car is only worth $25,000. Let \( t \) = number of years since the car was sold, and let \( V(t) \) = value of the car (in $).
2. **Questions:**
a) **Linear Function Determination:**
- Find a **linear function** for \( V(t) \) that fits the above data.
b) **Value of the Car After 5 Years:**
- What will the value of the car be 5 years after it was sold?
c) **Determining Year for a Specific Value:**
- During which year will the car’s value be at $8,000?
#### Detailed Breakdown:
**a) Finding a Linear Function:**
To find a linear function \( V(t) \), we will use the information given:
- Initial value (at \( t = 0 \)): \( V(0) = 41000 \)
- Value after 3 years (at \( t = 3 \)): \( V(3) = 25000 \)
The general form of a linear function is \( V(t) = V_0 - kt \), where:
- \( V_0 \) is the initial value
- \( k \) is the rate of depreciation per year
We can calculate \( k \) as follows:
\[ k = \frac{V_0 - V(3)}{3 - 0} = \frac{41000 - 25000}{3} = \frac{16000}{3} = 5333.33 \]
So the linear function \( V(t) \) becomes:
\[ V(t) = 41000 - 5333.33t \]
**b) Value of the Car After 5 Years:**
To find the value of the car after 5 years, substitute \( t = 5 \) into the linear function:
\[ V(5) = 41000 - 5333.33 \cdot 5 \]
\[ V(5) = 41000 - 26666.65 \]
\[ V(5) = 14333.35 \]
Therefore, the
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