In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years. The following information is available for car A - its value when new is £20000 - its value after one year is £16000 (a) Use an exponential model to form, for car A, a possible equation linking $V_t$ with $t$. (4) The value of car A is monitored over a 10-year period. Its value after 10 years is £2000 (b) Evaluate the reliability of your model in light of this information. (2) The following information is available for car B - it has the same value, when new, as car A - its value depreciates more slowly than that of car A (c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B. (1)

A-Level Edexcel Maths Pure Exponential & Logarithms: In a simple model, the value, $V

In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2

Question 9

In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years. The following information is available for car A - its value when new is £20000 - ... show full transcript

Worked Solution & Example Answer:In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2

Step 1

Use an exponential model to form, for car A, a possible equation linking $V_t$ with $t$.

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Answer

To model the depreciation of car A, we start with the exponential decay formula:

$V_t = V_0 e^{-kt}$

Where:

$V_0$ is the initial value (£20000)
$V_t$ is the value after $t$ years
$k$ is the decay constant.

Using the information that the value after one year is £16000:

Substitute $t = 1$ and $V_t = 16000$ :

$16000 = 20000 e^{-k}$
Solve for $k$ :

$e^{-k} = rac{16000}{20000} = 0.8$

Taking natural logarithm:

$-k = ext{ln}(0.8)$

Thus, $k ext{ is approximately } 0.223$ .
Finally, the model becomes:

$V_t = 20000 e^{-0.223t}$

Step 2

Evaluate the reliability of your model in light of this information.

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Answer

After 10 years, the model predicts:

Substitute $t = 10$ into the model:

$V_{10} = 20000 e^{-0.223 imes 10}$
Calculate:

$V_{10} ext{ is approximately } 2150.34$

This is significantly higher than the observed value of £2000.

Therefore, while the model provides a close estimate, it may not be entirely reliable, as it underestimates the depreciation in value over a longer period.

Step 3

Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

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Answer

To adjust the equation for car B, which depreciates more slowly than car A, we can modify the decay constant $k$ to a lower value.

For example, if car B depreciates more gently, we could set $k$ to a value like $0.18$ or $0.15$ to reflect reduced depreciation.

The new model would take the form:

$V_t = 20000 e^{-kt}$

with a suitable $k < 0.223$ .

In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2

Worked Solution & Example Answer:In a simple model, the value, $V_t$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 2

Use an exponential model to form, for car A, a possible equation linking $V_t$ with $t$.

Evaluate the reliability of your model in light of this information.

Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

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