Question

A point moves in the xy plane according to the law x=a sin $\omega$t, y=a(1-cos $\omega$t), where a and $\omega$ are positive constants.
(a)Find the distance s traversed by the point during the time t;
(b) Find the equation of trajectory.

• A. Distance=; trajectory =
• B. Distance =; Trajectory =
• C. Distance cannot be found, Trajectory =

Answer 1

The correct option is B

Distance =; Trajectory =

Okay so we have the position on the x and y plane as
x=a sin $\omega$t y=a(1-cos $\omega$t)
We can find the velocity in the x & y direction by differentiation
$v_x=\frac{dx}{dt}=a\omega cos\omega t;v_y=\frac{dy}{dt}=a\omega sin\omega t$
Now, speed is the magnitude of velocity.
So, $|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$
$=\sqrt{a^2{\omega }^{2}co{s}^2\omega t+a^2{\omega }^{2}si{n}^2\omega t}$
$|\overrightarrow{v}|=a\omega$
Oh wow! The speed is constant. It's not dependent on t. so I don't care what path the particle followed as I know that the particle travelled with a constant speed on the given path.
Oh great, now it's too simple $\Rightarrow$ distance = speed $\times$ time = a $\omega \times t$

Now to find how trajectory looks like, in other words how will the shape of the path look like on x, y plane. So basically I need to find the relation between y and x.
Hmmmmm.....
I have 2 equations. I just need to substitute something from equation (ii) to equation (i) to get this relation or.....
Wait
I know math, specifically trigonometry
I know $si{n}^2\theta +co{s}^2\theta =1$
I see them here
x=a sin $\omega$t, y=a(1-cos $\omega$t)
sin $\omega t=\frac{x}{a};cos\omega t=(1-\frac{y}{a})$
Now
$si{n}^2\omega t+co{s}^2\omega t=1$

$\Rightarrow \frac{x^2}{a^2}+(1-\frac{y}{a}{)}^2=1$
$\Rightarrow \frac{x^2}{a^2}+1+\frac{y^2}{a^2}-\frac{2y}{a}=1$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{a^2}-\frac{2y}{a}=0$
Equation of trajectory