The acceleration of the particle is defined by the relation \(a = 25 - 3x^2\ \text{mm/s}^2\). The particle starts with no initial velocity at the position x = 0. Determine the position where the velocity is maximum and the corresponding maximum veloc…

Q.1 (e) The acceleration of the particle is defined by the relation \(a = 25 - 3x^2\ \text{mm/s}^2\). The particle starts with no initial velocity at the position x = 0. Determine the position where the velocity is maximum and the corresponding maximum velocity.

[The exponent of x is partly obscured in the source scan; it is read as \(x^2\).]

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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Using \(v\dfrac{dv}{dx} = a\): \(\displaystyle\int_0^v v\,dv = \int_0^x (25 - 3x^2)\,dx \Rightarrow \dfrac{v^2}{2} = 25x - x^3\)

Maximum velocity when \(\dfrac{dv}{dx} = 0\), i.e. when \(a = 0\): \(25 - 3x^2 = 0 \Rightarrow x = \sqrt{25/3} = 2.887\) mm

\(v_{max}^2 = 2(25 \times 2.887 - 2.887^3) = 2(72.17 - 24.06) = 96.23\)

Answer: velocity is maximum at x = 2.887 mm and \(v_{max} = 9.81\) mm/s.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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The acceleration of the particle is defined by the relation \(a = 25 - 3x^2\ \text{mm/s}^2\). The particle starts with no initial velocity at the position x = 0. Determine the position where the velocity is maximum and the corresponding maximum veloc…

1 answer

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