Question

$x=4t^2,y=3t^2$ are the coordinates of a particle travelling in the$XY$plane at any time$\mathrm{t}$. At that point in time, the particle's speed is.

• A. $10t$
• B. $5t$
• C. $3t$
• D. $2t$
• E. $\sqrt{13t}$

Answer 1

The correct option is A

$10t$

Step 1: Given data:

The coordinates of a particle are,

$$\begin{gathered} x=4t^2 \\ y=3t^2 \end{gathered}$$

Step 2: Formula used:

As we know that velocity in x-direction for any particle is given as-

$V_x=\frac{dx}{dt}$

Velocity in y-direction-

$V_y=\frac{dy}{dt}$

Step 3: Calculating the velocity in x-direction

Using the formula, velocity in x-direction-

$\begin{array}{l}{V}_x=\left(\frac{dx}{dt}\right)\\ V_x=\frac{d\left(4t^2\right)}{dt}\\ V_x=8t\end{array}$

Step 4: Calculating the velocity in y-direction

Using the formula, velocity in x-direction-

$\begin{array}{l}\Rightarrow V_y=\left(\frac{dy}{dt}\right)\\ \Rightarrow V_y=\frac{d\left(3t^2\right)}{dt}\\ \Rightarrow V_y=6t\end{array}$

Step 5: At any time, the particle's velocity

$$\begin{gathered} \Rightarrow V=\sqrt{{V_x}^2+{V_y}^2} \\ \Rightarrow V=\sqrt{{\left(8t\right)}^2+{\left(6t\right)}^2} \\ \Rightarrow V=10t \end{gathered}$$

Hence, option A is the correct answer.