Question

The coordinates of a particle moving in $XY-$plane at any instant of time $t$ are$x=4t^2,y=3t^2$. The speed of the particle at that instant 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 moving in $XY-$plane at any instant of time $t$ are$x=4t^2,y=3t^2$.

Step 2: Formula used:

The speed of the particle $V=\sqrt{{V_x}^2+{V_Y}^2}$ here, $V_x$ the velocity of $X$and $V_Y$velocity of $Y$

Step 3: The speed of the particle

Note that $x=4t^2$differentiate with respect to $t$.

$$\begin{gathered} x=4t^2 \\ {V}_x=\frac{dx}{dt} \\ {V}_x=8t \end{gathered}$$

Similarly, differentiate $y=3t^2$

Therefore,

$V_Y=6t$

Now substitute the values in the speed formula,

The speed of the particle

$$\begin{gathered} V=\sqrt{{V_x}^2+{V_Y}^2} \\ V=\sqrt{{\left(8t\right)}^2+{\left(6t\right)}^2} \\ V=\sqrt{64t^2+36t^2} \\ V=10t \end{gathered}$$

Therefore the speed of the particle is $10t$.

Hence, the correct option is A.