Question

The equation of motion of a particle $P(x,y)$ on a plane are given by
$x=4+b\cos t,y=5+b\sin t$. Its velocity at time ${}^{\prime }{t}^{\prime }$ is

• A. $4$
• B. $5$
• C. $b$
• D. $\tan t$

Answer 1

Answer: (C)

$b$

We have,

$x=4+b\cos t$ ……. (1)

$y=5+b\sin t$ …….. (2)

On differentiating both equations with respect to $t$, we get

$\frac{dx}{dt}=0-b\sin t$

$V_x=-b\sin t$

$\frac{dy}{dt}=V_y=0+b\cos t$

$V_y=b\cos t$

Hence, the velocity

$V=\sqrt{{V_x}^2+{V_y}^2}$

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

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

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

We know that

${\cos }^{2}x+{\sin }^{2}x=1$

Therefore,

$V=\sqrt{b^2\times 1}$

$V=b$

Hence, this is the answer.