Multiple choice

When a $1700 kg$ car skids to a stop with initial speed $9 {m}/{s}$ on a flat road, it takes about $12 m$ to stop. Given that friction is the force allowing the car to decelerate, how much distance would the same car need to stop if it were initially moving at $15 {m}/{s}$?

  1. $33 m$
  2. $48 m$
  3. $20 m$
  4. $15 m$
  5. $18 m$
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A Correct answer
Explanation

Work-energy theorem states that the work done by friction equals the change in kinetic energy. F*d = 0.5*m*v^2. Since F is constant, d is proportional to v^2. d2/d1 = (v2/v1)^2 = (15/9)^2 = (5/3)^2 = 25/9. d2 = 12 * (25/9) = 33.33 m.

AI explanation

Using the kinematic equation v squared equals u squared plus 2as, the deceleration is constant for the same car on the same road. This means the stopping distance s is directly proportional to the square of the initial velocity u. The new distance is the original distance of 12 m multiplied by the ratio of the squares of the new and old speeds, which is 15 m/s divided by 9 m/s squared. Calculating 12 multiplied by 225 divided by 81 gives a stopping distance of approximately 33 m.

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