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    The question:
    The new price of a particular make of car is £10000
    When it is t years, the list price is £V. When t = 5, V = 5000

    The rate of depreciation is constant. Write this as a differential equation.
    Solve the differential equation and hence find the value of a car that is 7 years old according to this model.

    Explain why this model breaks down for large values of t.

    My response:
    • When t = 0 V = 10000
    • V = kt
    • 5000 = 5k
    • k = 1000?


    -dv/dt = kt
    -dv/dt = 1000 x 7
    dv/dt =-7000

    So therefore the value has decreased by £7,000 in 7 years, meaning the model would break down if t > 10 because the depreciation would be more than the original price of the car?

    I don't feel that I have answered this question correctly as I struggle to change the wording into the equation, any help would be greatly appreciated!
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    (Original post by Elucia)
    I don't feel that I have answered this question correctly as I struggle to change the wording into the equation, any help would be greatly appreciated!
    You were told that the rate of depreciation was constant

    So \dfrac{dV}{dt} = -k

    You need to solve this by doing

    \int dV = \int -k dt

    Then use the 2 pieces of information that you have to find c and k
 
 
 
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