Question

{"given":"Initial speed $u_1 = 50$ km/hr, stopping distance $s_1 = 6$ m. We need to find stopping distance $s_2$ when initial speed $u_2 = 100$ km/hr. The car uses the same braking system, so acceleration remains constant.","key_observation":"For motion with constant acceleration, the kinematic equation $v^2 = u^2 + 2as$ applies. At stopping, final velocity $v = 0$, so $s = \\frac{u^2}{2a}$. Since the braking acceleration is constant for the same car, the stopping distance is proportional to the square of the initial velocity.","option_analysis":[{"label":"(A)","text":"12 m","verdict":"incorrect","explanation":"This would be correct if stopping distance were directly proportional to velocity, but it's proportional to velocity squared. Since 100 km/hr is twice 50 km/hr, this gives only 2 times the original distance instead of 4 times."},{"label":"(B)","text":"18 m","verdict":"incorrect","explanation":"This value doesn't follow the correct proportionality relationship. The stopping distance should be proportional to the square of velocity, so it should be 4 times the original distance, not 3 times."},{"label":"(C)","text":"24 m","verdict":"correct","explanation":"Step 1: From kinematics, stopping distance $s = \\frac{u^2}{2a}$\n$$\\frac{s_2}{s_1} = \\frac{u_2^2}{u_1^2}$$\nStep 2: Substituting values:\n$$\\frac{s_2}{6} = \\frac{(100)^2}{(50)^2} = \\frac{10000}{2500} = 4$$\nStep 3: Therefore:\n$$s_2 = 4 \\times 6 = 24 \\text{ m}$$"},{"label":"(D)","text":"6 m","verdict":"incorrect","explanation":"This assumes the stopping distance remains the same regardless of speed, which is physically incorrect. Higher speeds require greater distances to stop due to higher kinetic energy that must be dissipated."}],"answer":"(C)","formula_steps":[]}