Question

The coordinates of a moving point particle in a plane at time $t$ is given by $x=a(t+\sin t),y=a(1-\cos t).$ The magnitude of acceleration of the particle is

• A. $a$
• B. $\sqrt{3}a$
• C. $2a$
• D. $\frac{\sqrt{3}}{2}a$

Answer 1

The correct option is A $a$

The magnitude of the acceleration along the $x$ direction
$a_x=\frac{d{x}^2}{d{t}^2}=\frac{d}{dt}\left(a\right(1+\cos t\left)\right)=-a\sin t$
The magnitude of the acceleration along the y direction
$a_y=\frac{d{y}^2}{d{t}^2}=\frac{d}{dt}\left(a\sin t\right)=a\cos t$
The resultant acceleration
$\sqrt{a_x^2+a_y^2}\Rightarrow \sqrt{(-a\sin t{)}^2+(a\cos t{)}^2}=a$