KINEMATICS

6 )

A particle moves from A to B and then it moves from B to C as shown in figure. Calculate the ratio between path length and displacement.

âˆž

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Answer :

Path length = d(AB) + d(BC) = 10 + 10 = 20 m

displacement = d(AC) = 10 m

ratio = (20 / 10) = 2

7 )

A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and displacement.

(2â„“ / âˆš3), â„“

(â„“ / âˆš3), â„“

2â„“, â„“

â„“, (2â„“ / âˆš3)

View Solution

Answer :

Draw PE ⊥ AB

PE bisects AB

m ∠ APB = 120°

in Δ PEB, ∠ EPB = 60°,

sin 60 = (EB / PB)

(√3 / 2) = {(ℓ/2) / (PB)}

PB = (ℓ / √3)

also PB = PC = PA = PD = (ℓ / √3)

Path length = d(AP) + d(PB) = (ℓ / √3) + (ℓ / √3) = (2 ℓ / √3)

displacement = d(AB) = ℓ

8 )

A Car goes from one end to the other end of a semicircular path of diameter 'd'. Find the ratio between path length and displacement.

(3Ï€ / 2)

Ï€

Ï€/2

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Answer :

{(path length) / (displacement)} = [{curve AB} / {d(AB)}]

= [{πd / 2} / d] = (π/2)

[half circumference = (πd / 2)]

9 )

A particle goes from point A to B. Its displacement is X and path length is y. So x/y ….

â‰¥ 1

â‰¤ 1

> 1

< 1

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Answer :

Path length is actual distance travelled.

Displacement is distance between two positions

hence path length ≥ displacement

y ≥ x i.e 1 ≥ (x/y)

(x/y) ≤ 1

10 )

As shown in the figure a particle starts its motion from 0 to A. And then it moves from A to B. AB is an arc find the Path length

r {1 + (Ï€/3)}

(Ï€/3) (r + 1)

r + (Ï€/3)

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Answer :

path length = d(OA) + d(AB) = r + arc AB

arc length = radius × θ

arc AB = rθ = r(π/3)

path length = r + (rπ / 3) = r{1 + (π/3)}

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