Let's analyze the problem step by step and provide solutions for each part.

Problem Breakdown:

1. Question (a):
   Check if the motorist will be arrested.

   • The motorist traveled 90 km at 60 km/hr, and then for $3.5$ hours at 80 km/hr.
   • Motorists are arrested if they exceed $80 \, \text{km/hr}$.

2. Question (b):

   • Calculate the stopping distance ($d$) for speeds of $46 \, \text{km/hr}$ and $85 \, \text{km/hr}$.
   • Find the speed ($s$) when the stopping distance is $35 \, \text{m}$.

The formula for stopping distance is given as: $d = \frac{s^2}{200} + \frac{s}{10}$

Part (a): Will the motorist be arrested?

Step 1: Time for each segment

• For the first $90 \, \text{km}$ at $60 \, \text{km/hr}$: $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{90}{60} = 1.5 \, \text{hours}$

• For the next $3.5 \, \text{hours}$ at $80 \, \text{km/hr}$: $\text{Distance covered} = \text{speed} \times \text{time} = 80 \times 3.5 = 280 \, \text{km}$

Step 2: Average speed

Total distance traveled: $\text{Total distance} = 90 + 280 = 370 \, \text{km}$

Total time taken: $\text{Total time} = 1.5 + 3.5 = 5 \, \text{hours}$

Average speed: $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{370}{5} = 74 \, \text{km/hr}$

Conclusion: The motorist will not be arrested, as their average speed ($74 \, \text{km/hr}$) does not exceed $80 \, \text{km/hr}$.

Part (b): Stopping Distance Calculations

Stopping distance formula:

$d = \frac{s^2}{200} + \frac{s}{10}$

(i) For $s = 46 \, \text{km/hr}$:

$d = \frac{46^2}{200} + \frac{46}{10} = \frac{2116}{200} + 4.6 = 10.58 + 4.6 = 15.18 \, \text{m}$

(ii) For $s = 85 \, \text{km/hr}$:

$d = \frac{85^2}{200} + \frac{85}{10} = \frac{7225}{200} + 8.5 = 36.125 + 8.5 = 44.625 \, \text{m}$

(iii) For $d = 35 \, \text{m}$, find $s$:

Use the quadratic equation: $d = \frac{s^2}{200} + \frac{s}{10}$ $35 = \frac{s^2}{200} + \frac{s}{10}$

Multiply through by $200$: $7000 = s^2 + 20s$

Rearranging: $s^2 + 20s - 7000 = 0$

Solve the quadratic equation: $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, \, b = 20, \, c = -7000$ $s = \frac{-20 \pm \sqrt{20^2 - 4(1)(-7000)}}{2(1)} = \frac{-20 \pm \sqrt{400 + 28000}}{2}$ $s = \frac{-20 \pm \sqrt{28400}}{2} = \frac{-20 \pm 168.54}{2}$

Taking the positive root: $s = \frac{-20 + 168.54}{2} = \frac{148.54}{2} = 74.27 \, \text{km/hr}$

Final Answers:

1. (a) The motorist will not be arrested.
2. (b) Stopping distances:
   • At $46 \, \text{km/hr}$: $15.18 \, \text{m}$
   • At $85 \, \text{km/hr}$: $44.63 \, \text{m}$
   • Speed for $35 \, \text{m}$: $74.27 \, \text{km/hr}$

Would you like detailed steps for any part?
Here are 5 related questions for further exploration:

1. What would be the stopping distance for $s = 100 \, \text{km/hr}$?
2. How does the stopping distance formula change if friction is considered?
3. Can we calculate the time taken to stop for a given distance?
4. What is the maximum stopping distance for $s = 120 \, \text{km/hr}$?
5. How does increasing $s$ affect $d$ (quadratic relationship)?

Tip: To solve equations involving both speed and distance, always check for unit consistency (e.g., meters vs kilometers).