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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

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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:

Vt=V0ektV_t = V_0 e^{-kt}

Where:

  • V0V_0 is the initial value (£20000)
  • VtV_t is the value after tt years
  • kk is the decay constant.

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

  1. Substitute t=1t = 1 and Vt=16000V_t = 16000:

    16000=20000ek16000 = 20000 e^{-k}

  2. Solve for kk:

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

    Taking natural logarithm:

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

    Thus, kextisapproximately0.223k ext{ is approximately } 0.223.

  3. Finally, the model becomes:

    Vt=20000e0.223tV_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:

  1. Substitute t=10t = 10 into the model:

    V10=20000e0.223imes10V_{10} = 20000 e^{-0.223 imes 10}

  2. Calculate:

    V10extisapproximately2150.34V_{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 kk to a lower value.

For example, if car B depreciates more gently, we could set kk to a value like 0.180.18 or 0.150.15 to reflect reduced depreciation.

The new model would take the form:

Vt=20000ektV_t = 20000 e^{-kt}

with a suitable k<0.223k < 0.223.

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