Question

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

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

Hint:

Recognize the (6, 8, 10) Pythagorean triplet to save time on the square root calculation in competitive exams!

Answer 1

The Correct Option is A

Concept: The velocity of a particle in a plane is a vector found by differentiating its position coordinates with respect to time. The speed is the magnitude of this velocity vector.
• Velocity Components: $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$.
• Speed ($v$): $v = \sqrt{v_x^2 + v_y^2}$ 171].

Step 1: Differentiate the coordinates.
Given $x = 4t^2$: \[ v_x = \frac{d}{dt}(4t^2) = 8t \] Given $y = 3t^2$: \[ v_y = \frac{d}{dt}(3t^2) = 6t \]

Step 2: Calculate the speed.
\[ v = \sqrt{(8t)^2 + (6t)^2} \] \[ v = \sqrt{64t^2 + 36t^2} \] \[ v = \sqrt{100t^2} \] \[ v = 10t \]