Multiple choice

A vehicle moving with speed $30\ km/h$ on a straight road can be stopped in $6\ m$ distance by applying brakes. If same vehicle is moving with $60\ km/h$ and brakes provide double retardation then vehicle will stop after travelling :

  1. $6\ m$
  2. $12\ m$
  3. $18\ m$
  4. $24\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using v^2 = u^2 + 2as. For the first case: 0 = (30)^2 + 2*a*6, so a = -900/12 = -75. For the second case: v=60, a' = 2a = -150. 0 = (60)^2 + 2*(-150)*s. 0 = 3600 - 300s, s = 12 m.

AI explanation

Using the third equation of motion, v^2 = u^2 + 2as, the stopping distance is given by s = u^2 / (2a). Initially, the distance is 6 = 30^2 / (2a), so 2a = 900 / 6 = 150. In the second scenario, the initial speed doubles to 60 km/h and the retardation doubles to 2a. The new stopping distance is s = 60^2 / (2 * 2a) = 3600 / 300, which equals 12 m.

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