Question

Hey there

with minute hand length
49cm
find distance and displacement.
if minute hand on 10 minutes​

Answer 1

In 60 minutes angle made by minute hand is 360

In 10 min angle = 360× 10/60 = 6×10= 60 degree

As arc length( ASB)= angle × radius(AO)

arc length would be distance

So distance = pie/3. × 49

= 22 ×49/7× 3

= 154/3 = 51.33 cm

Now displacement is shortest distance

So it would be AB

In AOB

AO = OB

and AOB = 60

So it becomes equilateral triangle

So AB= AO = 49 cm

So displacement = 49 cm

Answer 2

${\huge {\mathfrak {Answer :-}}}$

$\bold {Given :}$ There is a clock with a minute hand of length 49 cm.

$\bold {To \: find :}$ Distance and displacement if the minute hand covers 10 minutes.

____________________

⚠️ In 10 minutes, the minute hand moves from 12 to 2.

Look,

⚠️ In 60 mins, the minute hand makes 360° angle.

⚠️ In 1 min, the minute hand makes

= 360/60 = 6° angle.

⚠️ In 10 mins, the minute hand makes

= 10 × 6 = 60° angle.

✨ Hence, $\thita$ = 60°

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Distance is the total path covered by a body.

✨ DISTANCE

= arc AB

= (central angle)/360° × 2 × $\pi$ × r

= ${} 60°/360° × 2 × 22/7 × 49$

= ${} 1/6 × 2 × 22/7 × 49$

= ${} 1/3 × 22 × 7$

= ${} 154/3$

= 51.3

✨DISTANCE = 51.3 cm.

______________________

Displacement is the shortest distance covered by a body.

✨DISPLACEMENT = AB.

⚠️ In $\triangle$ AOB,

Since, AO and BO are the radius of the circle.

So, AO = BO.

⚠️ Since, the side opposite to equal angles are equal.

So, $\angle$ OAB = $\angle$ OBA.

⚠️ And, $\angle$ AOB = Central angle (thita) = 60°

⚠️ Now, by using the Anglesum property of a triangle,

$\angle$ OAB = (180° - 60°)/2

= 120°/2

= 60° = $\angle$ OBA

⚠️ Thus, in $\triangle$ AOB,

$\angle AOB = \angle OBA = \angle OAB$ = 60°

✨Hence, it is an equilateral triangle.

⚠️ The sides of an equilateral triangle are equal.

So, AB = AO = BO = 49 cm.

Hence, AB = 49 cm.

✨ DISPLACEMENT = 49cm.

______________________

◾DISTANCE = 51.3 cm.

◾DISPLACEMENT = 49 cm.

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✨HOPE THIS HELPS ✨