### Question StatementThe resultant force on a particle is given as \(\sum \mathbf{F} = 2t^3 \mathbf{i} + (3t^2 - 1) \mathbf{j} + (5t - t^2) \mathbf{k} \) Newton where \( t \) is in seconds. If the mass of the particle is \( m = 1.5 \, \text{kg} \), what is the magnitude of the acceleration of the particle at \( t = 1 \, \text{s} \)?### Options:- \( \bigcirc \) 3.27 \( \text{m/s}^2 \)- \( \bigcirc \) 4.90 \( \text{m/s}^2 \)- \( \bigcirc \) 7.35 \( \text{m/s}^2 \)- \( \bigcirc \) 1.5 \( \text{m/s}^2 \)### Explanation:To solve for the magnitude of acceleration, follow these steps:1. **Determine the Force at \( t = 1 \, \text{s} \):** Substitute \( t = 1 \) in the force equation, \(\sum \mathbf{F} = 2t^3 \mathbf{i} + (3t^2 - 1) \mathbf{j} + (5t - t^2) \mathbf{k} \). \[ \sum \mathbf{F} = 2(1)^3 \mathbf{i} + (3(1)^2 - 1) \mathbf{j} + (5(1) - (1)^2) \mathbf{k} \] Simplify each component: \[ \sum \mathbf{F} = 2 \mathbf{i} + (3 - 1) \mathbf{j} + (5 - 1) \mathbf{k} \] \[ \sum \mathbf{F} = 2 \mathbf{i} + 2 \mathbf{j} + 4 \mathbf{k} \, \text{N} \]2. **Calculate the Acceleration:** Using Newton's second law, \(\sum \mathbf{F} = m \mathbf{a}\), solve for \(\mathbf{a}\): \

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The resultant force on a particle is given as EF = 2t³i + (3t² − 1)j + (5t - t²)k Newton where t is in seconds. If the mass of the particle is m = 1.5 kg, what is the magnitude of the acceleration of the particle at t = 1 s? 3.27 m/s² 4.90 m/s² O 7.35 m/s² O 1.5 m/s²

The resultant force on a particle is given as EF = 2t³i + (3t² − 1)j + (5t - t²)k Newton where t is in seconds. If the mass of the particle is m = 1.5 kg, what is the magnitude of the acceleration of the particle at t = 1 s? 3.27 m/s² 4.90 m/s² O 7.35 m/s² O 1.5 m/s²

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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