The value of Bob’s car can be calculated from the formula $$V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500$$ where $V$ is the value of the car in pounds (£) and $t$ is the age in years - Edexcel - A-Level Maths Pure - Question 22 - 2013 - Paper 1

To find $t$ when $V = 9500$, we set up the equation:

$9500 = 17000e^{-0.25t} + 2000e^{-0.5t} + 500.$
Subtract 500 from both sides:

$9000 = 17000e^{-0.25t} + 2000e^{-0.5t}.$
Rearranging gives:

$17000e^{-0.25t} + 2000e^{-0.5t} - 9000 = 0.$
This is a transcendental equation that can be solved numerically or graphically to find:

$t ext{ approximately is } 4 ext{ (more precise calculations may depend on numerical methods)}.$

To find the rate of decrease of the car's value, we differentiate the value function:

rac{dV}{dt} = rac{d}{dt} (17000e^{-0.25t} + 2000e^{-0.5t} + 500).
Using the chain rule, we find:

rac{dV}{dt} = -4250e^{-0.25t} - 1000e^{-0.5t}.
Now we substitute $t = 8$ into the derivative:

rac{dV}{dt} igg|_{t=8} = -4250e^{-0.25(8)} - 1000e^{-0.5(8)}.
Calculating:

$e^{-0.25(8)} ext{ and } e^{-0.5(8)} ext{ and substituting them in gives a decrease of approximately £593 per year, to the nearest pound.}$