Kinematics in 2D Plane – Rankers Physics

Calculus Based Questions: Practice Problem & Solution

A particle is moving in x-y plane such that its x and y coordinates changes with time according to relation, \(x = 3t^2\) & \(y = 5t\) (here x & y are in m & t is in s). Speed of the particle at \(t = 2\) s, will be
17 \(\text{m s}^{-1}\)
\(\sqrt{34}\text{ m s}^{-1}\)
13 \(\text{m s}^{-1}\)
11 \(\text{m s}^{-1}\)

Solution Explained:

To solve this problem, we apply the core principles of Calculus Based Questions. Understanding the underlying formula is key to arriving at the correct answer below:

The velocity components are \(v_x = \frac{dx}{dt} = 6t\) and \(v_y = \frac{dy}{dt} = 5\). At \(t = 2\) s, \(v_x = 12\text{ m/s}\) and \(v_y = 5\text{ m/s}\). Speed is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 5^2} = 13\text{ m/s}\).

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