Question

The co-ordinates of a moving particle are $\mathrm{x}={\mathrm{at}}^2$ and $\mathrm{y}={\mathrm{bt}}^2$, where $\mathrm{a}$ and $\mathrm{b}$ are constants. the velocity of a particle at any instant is

• A. $2\mathrm{t}\sqrt{{\mathrm{a}}^2}++{\mathrm{b}}^2$
• B. $2\mathrm{t}\left(\mathrm{a}+\mathrm{b}\right)$
• C. $2\mathrm{t}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$
• D. $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Answer 1

The correct option is C

$2\mathrm{t}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Step 1. Given data

$\mathrm{x}={\mathrm{at}}^2$,

$\mathrm{y}={\mathrm{bt}}^2$

Step 2. Formula to be used.

Particle velocity is the velocity of a particle in a medium as it transmits a wave is called as the velocity of the particle.

The velocity of particle is,

$\mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}$

Step 3. Find the velocity of the particle.

The velocity of the particle is,

${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}}$

$=2\mathrm{at}$

${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}$

$=2\mathrm{bt}$

So,

$\mathrm{v}=\sqrt{4{\mathrm{a}}^2{\mathrm{t}}^2+4{\mathrm{b}}^2{\mathrm{t}}^2}$

$=2\mathrm{t}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

So, the velocity of the particle is $2\mathrm{t}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$.

Hence, option $\mathrm{C}$ is the correct answer.