Question

The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = a sin (pt)

where a, and p are positive contants of appropriate dimensions. Then,

• A. the equation of trajectory is $x^2+y^2=a^2$
• B. the velocity and acceleration of the particle are normal to each other
• C. the acceleration of the particle is always directed towards the center
• D. the distance travelled by the particle in time interval $(t=0tot=\frac{\pi }{2\rho }$) is "a”.

Answer 1

Answer: (A), (B), (C)

the acceleration of the particle is always directed towards the center

x(t) = a cos (pt) ; y(t) = a sin(pt)

Now to get an equation in x & y and to eliminate t

cos (pt) = $\frac{x}{a}$ & sin(pt) = $\frac{y}{a}$

$co{s}^2\left(pt\right)+si{n}^2\left(pt\right)$ = 1

$\Rightarrow$ $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$

$x^2+y^2=a^2$ $\therefore$ option (a) is correct

b) Lets find velocity and acceleration

$v_{\left(x\right)}=\frac{dx}{dt}=-ap$ sin(pt)

$v_y=\frac{dy}{dt}=ap$ cos pt

Slope of velocity = $\frac{v_y}{v_x}$= (-cot pt)

$a_x=\frac{d{v}_x}{dt}=-a{p}^2$ cos pt

$a_y=\frac{d{v}_y}{dt}=-a{p}^2$ sin pt

Slope of acceleration = $\frac{a_y}{a_x}$ = tan pt

Now if two lines are perpendicular then the product of their slopes is -1

Also since it's a uniform circular motion as ( magnitude of velocity $\sqrt{(apsinpt{)}^2+(apcospt{)}^2}$ = ap is

constant)

So acceleration will be perpendicular to velocity.

The position of the particle that is measured from the centre is given as x = a cos pt ; y = a sin pt

$\overrightarrow{r}$ = a ( cospt $\hat{i}$ + sin pt $\hat{j}$ )

Now acceleration is

$a_x=-a{p}^{2}cospt$ ; $a_y=-a{p}^{2}sinpt$

$\overrightarrow{a}=-a{p}^2(cospt\hat{i}+sinpt\hat{j}$ )

$\Rightarrow$ $\overrightarrow{a}=-p^2\overrightarrow{r}$

So basically the acceleration is in the opposite direction of radius.

And since radius starts from centre and points outwards so that means the acceleration points inwards the circle,

So option c is correct

Now option d asks about the distance travelled.

We agreed that whatever i want to find distance i will need speed which is magnitude of velocity.

Speed = $\left(\overrightarrow{v}\right)$ = $\sqrt{v_x^2+v_y^2}=\sqrt{(-apsinpt{)}^2+(apcospt{)}^2)}$

= ap which is constant

So s = $\frac{d}{t}$ d = st = ap × $\frac{\pi }{2p}$

So option d is wrong .