1. Given data

   • Initial speed of car: $u = 100\,\text{m/s}$
   • Final speed after stopping: $v = 0$
   • Coefficient of kinetic friction: $\mu_k = 0.5$

2. Retarding force and deceleration

   The maximum retarding acceleration due to friction is $a = \mu_k g$ acting opposite to motion.

   Taking $g \approx 10\,\text{m/s}^2$, $a = 0.5 \times 10 = 5\,\text{m/s}^2$

   Since it is retardation, $a = -5\,\text{m/s}^2$

3. Use kinematic equation

   We use $v^2 = u^2 + 2as$

   Substituting $v=0$, $u=100$, $a=-5$: $0 = (100)^2 + 2(-5)s$ $0 = 10000 - 10s$ $10s = 10000$ $s = 1000\,\text{m}$

4. Match with options

   • A: $1000\,\text{m}$ ✅
   • B: $800\,\text{m}$
   • C: $400\,\text{m}$
   • D: $100\,\text{m}$

Therefore, the stopping distance is $\boxed{1000\,\text{m}}$

5. Comparison with stored correct answer

   Stored correct answer: A

   Our derived answer is also A. So they agree.