Question

At any time \( t \), the co-ordinates of moving particle are \( x = at^2 \) and \( y = bt^2 \). The speed of the particle is

• A. \( 2t\sqrt{a^2 + b^2} \)
• B. \( 2t\sqrt{a^2 - b^2} \)
• C. \( 2t(a + b) \)
• D. \( \frac{2t}{\sqrt{a^2+b^2}} \)

Hint:

Speed is the magnitude of the velocity vector: $|v| = \sqrt{\dot{x}^2 + \dot{y}^2}$.

Answer 1

The Correct Option is A

Step 1: Differentiate to find velocity
$v_x = \frac{dx}{dt} = 2at$.
$v_y = \frac{dy}{dt} = 2bt$.
Step 2: Find speed
Speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(2at)^2 + (2bt)^2}$.
Step 3: Simplify
$v = \sqrt{4a^2t^2 + 4b^2t^2} = 2t\sqrt{a^2 + b^2}$.
Final Answer: (A)