A car stops at two sets of traffic lights - Edexcel - A-Level Maths Pure - Question 10 - 2022 - Paper 1

To determine when the maximum speed occurs, we differentiate the speed function:

$v = (10 - 0.4t) \ln(t + 1)$

Using the product rule:

$\frac{dv}{dt} = \left((10 - 0.4) \frac{1}{t+1} + \ln(t + 1)(-0.4)\right)$ Set ( \frac{dv}{dt} = 0 ) to find critical points:

1. This leads to: ( (10 - 0.4) \frac{1}{t + 1} = 0.4 \ln(t + 1) )
2. Rearranging gives: ( t^* = \frac{26}{1 + \ln(t + 1)} - 1 ) Thus proving the maximum speed occurs at ( t^*).

Using the iteration formula:

$t_{n+1} = 26 \div (1 + \ln(t_n + 1)) - 1$

1. Start with ( t_1 = 7 ): ( t_2 = 26 \div (1 + \ln(7 + 1)) - 1 \approx 7.298 )
2. Continue iterating: ( t_3 = 26 \div (1 + \ln(7.298 + 1)) - 1 \approx 7.33 )
3. Hence, to three decimal places, ( t_3 \approx 7.33 ) seconds.