Question

A particle is moving in a circular path of radius t.Calculate displacement & distance after half a circle

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Answer 1

Answer:

The answer will be 2r.

Explanation:

After the particle covers half path of the circle, the displacement can be easily calculated by measuring the distance between initial and final points. Displacement after half circle = AB=OA + OB Hence, the displacement after half circle is 2r.

Answer 2

Anѕwєr :

$\Large\boxed{\red{\sf{Distance = \pi t }}}$

$\Large\boxed{\red{\sf{Displacement = 2t }}}$

Explαnαtion :

According to the question, we have to calculate the displacement & distance after the particle covers the half a circle. Here,

• Radius,r = t

Now, have a look at the attachment. Suppose the particle starts from A. So, after covering half of the circular path it'll come to point B.

Let the distance covered be D and displacement be S.

The distance covered will be the the total length (Perimeter) of the semicircle formed from A to B. Perimeter of the semicircle is given by the half of the perimeter of the circle. So

$\implies{\sf{ D = \dfrac{Perimeter \; of\; path}{2} }}$

Perimeter of the circular path is 2πr.

$\implies{\sf{ D = \dfrac{2\pi r}{2} }}$

$\implies{\sf{ D = \dfrac{2\pi t}{2} }}$

$\implies\boxed{\red{\sf{ D = \pi t}}}$

∴ The distance covered by the particle after it covers half of the circle is πt.

Now, we have to calculate the displacement. Displacement is the shortest distance from initial to final position of the body. Here, the initial position is A and the final position is B. The shortest distance from A to B will be the line AB. AB is also the diameter of the circle. So,

$\implies{\sf{ S = Diameter}}$

$\implies{\sf{ S = 2 \times Radius}}$

$\implies\boxed{\red{\sf{ S = 2t}}}$

∴ The displacement of the the particle after it covers half of the circle is 2t.

$\rule{200}2$