Multiple choice

An ant moves with a uniform speed along a straight line path from P to Q. 60 cm apart, in front of a mirror and takes two minutes to cover this distance. The line segment PQ is inclined to the mirror $M{M}'$ at an angle of $30^{\circ}$. What is the rate at which the distance between the ant and its image changes?

  1. $0.5 cm s^{-1}$
  2. $1.5 cm s^{-1}$
  3. $1.0 cm s^{-1}$
  4. $2.5 cm s^{-1}$
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A Correct answer
Explanation

The ant moves at speed v = 60 cm / 120 s = 0.5 cm/s. The angle to the mirror is 30 degrees. The component of velocity perpendicular to the mirror is v * sin(30) = 0.5 * 0.5 = 0.25 cm/s. The image moves at the same speed in the opposite direction, so the relative speed between the ant and its image is 2 * 0.25 = 0.5 cm/s.

AI explanation

The ant travels the 60 cm distance in 120 seconds, giving a speed of 0.5 cm per second along the line. Since the mirror image moves symmetrically opposite to the ant, the distance between them closes at twice the ant's velocity component perpendicular to the mirror. The perpendicular component is found using sin 30 degrees, giving a rate of 2 times 0.5 times sin 30 degrees. This calculates to 1 times 0.5, which equals 0.5 cm per second.

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