11 The second hand of a watch is $15 \mathrm{~mm}$ long. How far does the pointer travel in (a) 15 seconds (b) 20 seconds (c) $5 \frac{1}{2}$ minutes? (Take $\pi=3.14$ )

Step 1 ：The second hand of a watch moves in a circular path. The distance travelled by the second hand is the length of the arc it sweeps out. The length of an arc is given by the formula: Arc length = radius * angle (in radians).

Step 2 ：The radius of the circle is the length of the second hand, which is 15mm. The angle is determined by the amount of time the second hand has moved. In one minute (or 60 seconds), the second hand makes a complete revolution, which is 2π radians. Therefore, the angle in radians for t seconds is (t/60)*2π.

Step 3 ：Let's calculate the distance travelled by the second hand for each of the given times.

Step 4 ：For 15 seconds, the distance travelled by the second hand is \(15 * \frac{15}{60} * 2 * 3.14\) which equals to \(23.56 \text{ mm}\).

Step 5 ：For 20 seconds, the distance travelled by the second hand is \(15 * \frac{20}{60} * 2 * 3.14\) which equals to \(31.42 \text{ mm}\).

Step 6 ：For $5 \frac{1}{2}$ minutes (or 330 seconds), the distance travelled by the second hand is \(15 * \frac{330}{60} * 2 * 3.14\) which equals to \(518.36 \text{ mm}\).

Step 7 ：Final Answer: (a) The distance travelled by the second hand in 15 seconds is \(\boxed{23.56 \text{ mm}}\). (b) The distance travelled by the second hand in 20 seconds is \(\boxed{31.42 \text{ mm}}\). (c) The distance travelled by the second hand in $5 \frac{1}{2}$ minutes is \(\boxed{518.36 \text{ mm}}\).