Question

The minute hand of a clock is 14m long. the approximate distance covered and displacement of the tip of the minute hand during the interval the time changes from 5:05 to 7:20 pm are =

Answer 1

Step 1: Given that:

The length of the minute hand of the clock= $14m$

Given time interval= $5:05pm-7:20pm$ = $2hr15minutes$

Step 2: The distance covered by the minute hand of the clock in given time:

1. The minute hand of a clock completes one round in $1hror60minutes$ .
2. Thus the minute hand will cover two complete rounds and in the next $15minutes$ , it will cover $\frac{1}{4}$ round of the clock.

Now,

The distance covered by the minute hand in two complete rounds is given by;

$d=2\times$ Circumference of the circular path covered by minute hand with its length as the radius of the path.

Thus,

$d=2\times 2\pi r$ , where r is the length of the minute hand of the clock

$d=2\times 2\times \frac{22}{7}\times 14m$

$d=4\times 22\times 2$

$d=196m$

The distance covered by the minute hand in the next 15 minutes will be

$d^{\prime }=\frac{1}{4}\times$ the circumference covered by the minute hand with its length as the radius

$d^{\prime }=\frac{1}{4}\times 2\times \frac{22}{7}\times 14m$

$d^{\prime }=11\times 2$

$d^{\prime }=22m$

Thus,

The total distance covered by the minute hand in the given time will be

$D=d+d^{\prime }$

$D=196m+22m$

$D=218m$

Step 3: Calculation of displacement of the tip of the minute hand in the given time:

In one complete round,

Displacement of minute hand= $0$

Thus, the displacement of the tip of the minute hand in 2hrs that is in two complete rounds will be $0$.

In the next $15minutes$ displacement will be:

Thus, the displacement will be calculated by using the Pythagoreus theorem

Displacement= $\sqrt{({14}^2+{14}^2)}$

= $14\sqrt{2}m$

Thus total displacement in the given time will be, $14\sqrt{2}m$ .

Hence,

In given time,

Total distance covered by minute hand= $218m$

Total displacement covered = $14\sqrt{2}m$