A point moves in the x-y plane according to x=Asinω t and y=A(1 - textcos ω t) where A and ω are positive constants. The distance traversed by the point during the time τ is

Question

A point moves in the x-y plane according to x=Asinω t and y=A(1 - textcos ω t) where A and ω are positive constants. The distance traversed by the point during the time τ is (A) (A ω 2/τ) (B) Aω 2τ (C) Aω τ (D) (A ω /τ)

Tardigrade Admin · Accepted Answer

Distance travelled by point in time t is:

S = \int_{0}^{τ} ∣ ∣ V \to ∣ ∣ d t

∵ ∣ ∣ V \to ∣ ∣ = V_{x}^{2} + V_{y}^{2}

Where,

V_{x} = \frac{d x}{d t} = A ω cos ω t an d

V_{y} = \frac{d y}{d t} = A ω s inω t

S = \int_{0}^{τ} A ω d t = A ω τ

Q. A point moves in the $x - y$ plane according to $x = A s inω t$ and $y = A (1 - cos ω t)$ where $A$ and $ω$ are positive constants. The distance traversed by the point during the time $τ$ is

$\frac{A ω ^{2}}{τ}$

$A ω^{2} τ$

$A ω τ$

$\frac{A ω}{τ}$

Distance travelled by point in time t is:
$S = \int_{0}^{τ} ∣ ∣ V \to ∣ ∣ d t$
$∵ ∣ ∣ V \to ∣ ∣ = V_{x}^{2} + V_{y}^{2}$
Where, $V_{x} = \frac{d x}{d t} = A ω cos ω t an d$ $V_{y} = \frac{d y}{d t} = A ω s inω t$
$S = \int_{0}^{τ} A ω d t = A ω τ$

Q. A point moves in the x−y plane according to x=Asinωt and y=A(1−cosωt) where A and ω are positive constants. The distance traversed by the point during the time τ is

τAω2​

Aω2τ

Aωτ

τAω​