1. **Problem statement:** A car moving at a speed of 126 km/h is brought to a stop within 200 m. We need to find (i) the retardation (deceleration) of the car and (ii) the time it takes to stop. 2. **Given data:** - Initial speed, $u = 126$ km/h - Final speed, $v = 0$ (since the car stops) - Distance, $s = 200$ m 3. **Convert speed to m/s:** $$u = 126 \times \frac{1000}{3600} = 35 \text{ m/s}$$ 4. **Formula for retardation:** Using the equation of motion: $$v^2 = u^2 + 2as$$ where $a$ is acceleration (retardation here, so it will be negative). Rearranged: $$a = \frac{v^2 - u^2}{2s}$$ 5. **Calculate retardation:** $$a = \frac{0^2 - 35^2}{2 \times 200} = \frac{-1225}{400} = -3.0625 \text{ m/s}^2$$ Retardation is $3.0625$ m/s$^2$ (negative sign indicates deceleration). 6. **Formula for time to stop:** Using: $$v = u + at$$ Rearranged: $$t = \frac{v - u}{a}$$ 7. **Calculate time:** $$t = \frac{0 - 35}{-3.0625} = \frac{-35}{-3.0625} = 11.43 \text{ seconds}$$ **Final answers:** (i) Retardation = $3.06$ m/s$^2$ (ii) Time to stop = $11.43$ seconds