Acceleration of a Particle from Coordinates – Rankers Physics
Topic: Kinematics
Subtopic: Calculus Based Questions

Acceleration of a Particle from Coordinates

The x and y co-ordinates of a particle at any time \(t\) are given by : \(x = 7t + 4t^2\) and \(y = 5t\) where \(x\) and \(y\) are in m and \(t\) in s. The acceleration of the particle at 5 s is :-
Zero
\( 8 m/s^{2}\)
\(20 m/s^{2}\)
\(40 m/s^{2} \)

Solution:

Velocity components: \(v_x = \frac{dx}{dt} = 7 + 8t\) and \(v_y = \frac{dy}{dt} = 5\). Acceleration components: \(a_x = \frac{dv_x}{dt} = 8\) and \(a_y = \frac{dv_y}{dt} = 0\). Total acceleration is \(a = \sqrt{a_x^2 + a_y^2} = 8\text{ m/s}^2\) at any time.

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