A point moves in the plane `xy` according to the law `x=a sin omegat`, `y=a(1-cos omega t)`, where a and `omega` are positive constants. Find:
(a) the distance s traversed by the point during the time `tau`,

Options

A. `(Aomega^2)/(tau)`
B. `A omega^2 tau`
C. `A omega tau`
D. `(Aomega)/(tau)`

Answer

To solve the problem, we need to find the distance \( s \) traversed by the point during the time \( \tau \) given the equations of motion:

\[ x = a \sin(\omega t) \]
\[ y = a(1 - \cos(\omega t)) \]

### Step 1: Find the velocity components

First, we need to find the velocity components \( v_x \) and \( v_y \) by differentiating the position equations with respect to time \( t \).

- For \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]

- For \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(a(1 - \cos(\omega t))) = a \omega \sin(\omega t) \]

### Step 2: Find the magnitude of the velocity

Now, we can find the magnitude of the velocity \( v \) using the Pythagorean theorem:

\[ v = \sqrt{v_x^2 + v_y^2} \]

Substituting the expressions for \( v_x \) and \( v_y \):

\[ v = \sqrt{(a \omega \cos(\omega t))^2 + (a \omega \sin(\omega t))^2} \]

This simplifies to:

\[ v = \sqrt{a^2 \omega^2 (\cos^2(\omega t) + \sin^2(\omega t))} \]

Using the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):

\[ v = \sqrt{a^2 \omega^2} = a \omega \]

### Step 3: Calculate the distance traveled

The distance \( s \) traveled by the point during the time \( \tau \) can be calculated by integrating the magnitude of the velocity over the time interval from \( 0 \) to \( \tau \):

\[ s = \int_0^{\tau} v \, dt = \int_0^{\tau} a \omega \, dt \]

Since \( a \omega \) is a constant, we can take it out of the integral:

\[ s = a \omega \int_0^{\tau} dt = a \omega [t]_0^{\tau} = a \omega (\tau - 0) = a \omega \tau \]

### Final Answer

Thus, the distance \( s \) traversed by the point during the time \( \tau \) is:

\[ s = a \omega \tau \]

---