Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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![**Question:**
A particle moves along a straight line such that its displacement at any time \( t \) is given by
\[ x = (t^3 - 6t^2 + 3t + 4) \, \text{m} \]
The velocity when the acceleration is zero is:
- \( 3 \, \text{ms}^{-1} \)
- \(-12 \, \text{ms}^{-1} \)
- \( 42 \, \text{ms}^{-1} \ )
- \(-9 \, \text{ms}^{-1} \)
**Solution Explanation:**
To determine the velocity when the acceleration is zero, we need to perform the following steps:
1. **First Derivative (Velocity):** Find the first derivative of the displacement function \( x(t) \) to get the velocity function \( v(t) \).
\[ v(t) = \frac{dx}{dt} = 3t^2 - 12t + 3 \]
2. **Second Derivative (Acceleration):** Find the second derivative of the displacement function \( x(t) \) to get the acceleration function \( a(t) \).
\[ a(t) = \frac{d^2x}{dt^2} = 6t - 12 \]
3. **Set Acceleration to Zero:** Set the acceleration function \( a(t) \) to zero and solve for \( t \).
\[ 0 = 6t - 12 \]
\[ t = 2 \]
4. **Calculate Velocity at \( t = 2 \):** Substitute \( t = 2 \) into the velocity function \( v(t) \).
\[ v(2) = 3(2)^2 - 12(2) + 3 \]
\[ v(2) = 3(4) - 24 + 3 \]
\[ v(2) = 12 - 24 + 3 \]
\[ v(2) = -9 \, \text{ms}^{-1} \]
Therefore, the velocity when the acceleration is zero is \(-9 \, \text{ms}^{-1} \).](https://content.bartleby.com/qna-images/question/fa64e205-1705-4602-bbb7-7e7367e5cb05/b9598bf1-6347-4c6f-aff0-b176b8001787/7r3rbct_processed.png)
Transcribed Image Text:**Question:**
A particle moves along a straight line such that its displacement at any time \( t \) is given by
\[ x = (t^3 - 6t^2 + 3t + 4) \, \text{m} \]
The velocity when the acceleration is zero is:
- \( 3 \, \text{ms}^{-1} \)
- \(-12 \, \text{ms}^{-1} \)
- \( 42 \, \text{ms}^{-1} \ )
- \(-9 \, \text{ms}^{-1} \)
**Solution Explanation:**
To determine the velocity when the acceleration is zero, we need to perform the following steps:
1. **First Derivative (Velocity):** Find the first derivative of the displacement function \( x(t) \) to get the velocity function \( v(t) \).
\[ v(t) = \frac{dx}{dt} = 3t^2 - 12t + 3 \]
2. **Second Derivative (Acceleration):** Find the second derivative of the displacement function \( x(t) \) to get the acceleration function \( a(t) \).
\[ a(t) = \frac{d^2x}{dt^2} = 6t - 12 \]
3. **Set Acceleration to Zero:** Set the acceleration function \( a(t) \) to zero and solve for \( t \).
\[ 0 = 6t - 12 \]
\[ t = 2 \]
4. **Calculate Velocity at \( t = 2 \):** Substitute \( t = 2 \) into the velocity function \( v(t) \).
\[ v(2) = 3(2)^2 - 12(2) + 3 \]
\[ v(2) = 3(4) - 24 + 3 \]
\[ v(2) = 12 - 24 + 3 \]
\[ v(2) = -9 \, \text{ms}^{-1} \]
Therefore, the velocity when the acceleration is zero is \(-9 \, \text{ms}^{-1} \).
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