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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![### Question Statement
The 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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Transcribed Image Text:### Question Statement
The 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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