Question

▶️In a clock the minute hand has radius 49 cm, so find distance and displacement if minute on 20 minutes

▪️▪️▪️ Answer!!!​

Answer 1

Angle made by minute hand in 60 min= 360

So angle made in 20 min = 360/3 = 120

as 60/3 = 20 So its 1/3 rd of 360

Or

360× 20/60 = 6× 20 = 120

Arc length ( ASB)( Distance ) = theta × radius

= 120 × pie/180 ( radian converted) × 49

= 120 × 22 × 49/(7× 180)

= 120× 22 × 7 / 180

= 2× 22× 7/ 3 = 308/3= 102.66

Displacement ( AB)

Draw perpendicular from O to AB

As its isosceles triangle

so it would be perpendicular bisector

also angle bisect

AOF = 60

In OAF

AF = sin60 × AO = 49 × √3/2

As AF = FB

So AB = 2× 49√3/2 = 49√3 which is displacement

Answer 2

Answer:

• Given radius = 49cm

The angle made by the minute hand in 20 min is = 120 °

{°•° Angle made by minute hand in 60 second is 360 ° so 20 min is ⅓ of 60 min }✓

Now:

_________

$length \: of \: arc = theta \times radius \: of \: circle$

•°• length of arc =

$120 \times \binom{\pi}{180} \times 49$

Note :-

$\frac{\pi}{180} = radian \: converted$

$= > 120 \times 22 \times \frac{49}{180 \times 7} \\ \\ = > \frac{308}{3} \\ \\ = > 102.6 \: cm$

102.6 cm is the distance traveled by the minute hand in 20 min ✓

Now let us find the displacement :

__________________________________________

In ∆OPT angle POT is besected by OS and PS=TS

So in ∆ OPS

angle O= 60° { °•° ½ of angle POT}

OP = 49 { radius of the circle }✓

$therefore \: = \frac{ps}{po} = sin60 \\ \\ = > ps = sin60 \times op \\ \\ = > ps = 49 \times \frac{ \sqrt{3} }{2} \\ \\ = > ps = \frac{49 \sqrt{3} }{2} \\ let\\ \sqrt{3} = 1.7 \\ \\ so \: ps = \frac{83.3}{2} = 41.65cm$

•°• PT or the displacement is = 2×PS

=> PT = 2 × 41.65

=> PT = 83.3 cm

$the \: displacement \: is \: = </strong><strong>8</strong><strong>3</strong><strong>.3cm$