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

Answer 1

Answer: (C)

$a$

Clearly, the position vector of particle at any time $t$ is
$r=a(t+\sin t)\hat{i}+a(1-\cos t)\hat{j}$

Now, $\frac{dr}{dt}=a(1+\cos t)\hat{i}+a\sin t\hat{j}$

$\therefore \frac{d^{2}r}{d{t}^2}=-a\sin t\hat{i}+a\cos t\hat{j}$

Hence, the required acceleration at any time $t$

$=-a\sin t\hat{i}+a\cos t\hat{j}$

and magnitude of the acceleration

$=\sqrt{{(-a\sin t)}^2+{\left(a\cos t\right)}^2}$

$=\sqrt{a^2{\sin }^{2}t+a^2{\cos }^{2}t}$

$=\sqrt{a^2({\sin }^{2}t+{\cos }^{2}t)}$

$=\sqrt{a^2\times 1}=a$.