Calculus Based Questions: Practice Problem & Solution
The motion of a particle along a straight line is described by equation: \\(x = 8 + 12t - t^3\) where (x) is in metre and (t) in second. The retardation of the particle when its velocity becomes zero, is: (2012 Pre)
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:
Given \(x = 8 + 12t - t^3\). Velocity \(v = \frac{dx}{dt} = 12 - 3t^2\). Acceleration \(a = \frac{dv}{dt} = -6t\). When (v=0), \(12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}\). At \(t=2 \text{ s}\), \(a = -6(2) = -12 \text{ m/s}^2\). Retardation is \(-a = 12 \text{ m/s}^2\).
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