Question

A person moves on a circular track of radius
R = 10 m in anticlockwise sense, along ACB
such that LAOB = 120°. Find distance
traversed and magnitude of displacement​

Answer 1

Distance is equal to the length of the circular arc of central angle 120°. Thus,

$\displaystyle\longrightarrow\sf {d=2\pi r\times\dfrac {120^{\circ}}{360^{\circ}}}$

$\displaystyle\longrightarrow\sf {d=2\pi\times 10\times\dfrac {1}{3}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {d=\dfrac {20\pi}{3}\ m}}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {d=20.94\ m}}}$

Displacement is equal to the length of the circular chord of central angle 120°, which can form a traingle with radii as the other two sides with 120° between them. Thus, by cosine rule,

$\displaystyle\longrightarrow\sf{s=\sqrt{r^2+r^2-2\times r\times r\cos 120^{\circ}}}$

$\displaystyle\longrightarrow\sf{s=\sqrt{2r^2-2r^2\cos 120^{\circ}}}$

$\displaystyle\longrightarrow\sf{s=\sqrt{2r^2(1-\cos 120^{\circ})}}$

$\displaystyle\longrightarrow\sf{s=\sqrt{2r^2\cdot2\sin^260^{\circ}}}$

$\displaystyle\longrightarrow\sf{s=2r\sin60^{\circ}}$

$\displaystyle\longrightarrow\sf{s=2\times10\times\dfrac {\sqrt3}{2}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {s=10\sqrt3\ m}}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {s=17.32\ m}}}$

Answer 2

The distance traversed is 20.93 m.

The magnitude of displacement is = 17.32 m.

Given: A person moves on a circular track of radius R = 10 m in an anticlockwise sense, along ACB such that LAOB = 120°.

To Find: The distance traversed and the magnitude of displacement​.

Solution:

• We know that the length of an arc can be calculated using the formula,

         Length of arc = R × Ф                                        ....(1)

Where, R = radius, Ф = angle subtended by the arc at the center in radian

• We know that to covert angle to radians, we make use of the formula,

        Ф° = ( Ф / 180 ) × π   radians                                ....(2)

where Ф = angle in degree.

Coming to the numerical, we are given;

The radius (R) of the track = 10 m, ∠ AOB (Ф) = 120°

Also, from (2), ∠ AOB =  ( 120 / 180 ) × π   radians  = 2π/3 radians

Putting respective values in (1), we get;

          Length of arc = R × Ф

     ⇒  Length of arc = 10 × 2π/3 = ( 10 × 2 × 3.14 ) / 3

                                  = 20.93 m

So, the distance traversed is 20.93 m.

We know that displacement is the shortest possible distance between two points A and B. So, we can visualize that the shortest distance, in this case, is the distance AB.

Now, in Δ AOB, AO = OB = 10 m and ∠ AOB = 120°.

So, the distance AB = 2 × R sin 60°

                                 = 2 × 10 × √3/2

                                 = 10√3 m

                                 = 17.32 m

]So, the magnitude of displacement is = 17.32 m.

Hence, the distance traversed is 20.93 m.

The magnitude of displacement is = 17.32 m.

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