A car stops at two sets of traffic lights - Edexcel - A-Level Maths Pure - Question 11 - 2022 - Paper 1

To find the maximum speed, we differentiate the speed function with respect to t:

$\frac{dv}{dt} = 0$

Using product and chain rules, we set:

$(10 - 0.4)\text{ln}(t + 1) + (0 - 0.4t \cdot \frac{1}{t + 1})(t) = 0$

This leads us to rearranging and solving:

1. Setting $\frac{dv}{dt} = 0$, we realize: $\frac{26}{1 + \text{ln}(t + 1)} - 1$

2. Follow through to find the conditions for $t$ that maximizes v. This yields: $t^* = \frac{26}{1 + \text{ln}(t + 1)} - 1$

Using the iteration formula:

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

Start with initial value $t_1 = 7$:

• For $n = 1$: $t_2 = \frac{26}{1 + \text{ln}(7 + 1)} - 1 \approx 7.298$
• Round as necessary for precision. Thus, after solving, to three decimal places, we find: $t_1 = 7.298$.

To estimate the maximum speed further, continue iteration using the last value:

• Replace $t_1 = 7.298$ into: $t_{n+1} = \frac{26}{1 + \text{ln}(t_n + 1)} - 1$
• Continue the iterations:
• After subsequent iterations, we find: $t_n \approx 7.33$ seconds as the converging value after continued calculations.