Acceleration vs. position of a particle is as given in the figure. If the velocity v = 5 m/s at s = 0, what is the velocity at s = 100 m? a [m/s²] 4 2 O24.49 m/s O 15 m/s O 10 m/s O 25 m/s +----18 100 s [m]
Acceleration vs. position of a particle is as given in the figure. If the velocity v = 5 m/s at s = 0, what is the velocity at s = 100 m? a [m/s²] 4 2 O24.49 m/s O 15 m/s O 10 m/s O 25 m/s +----18 100 s [m]
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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![**Acceleration vs. Position of a Particle**
The question provided is:
*Acceleration vs. position of a particle is as given in the figure. If the velocity \( v = 5 \text{ m/s} \) at \( s = 0 \), what is the velocity at \( s = 100 \text{ m}?*
The diagram shows the acceleration \( a \) of a particle in relation to its position \( s \).
- The horizontal axis represents position \( s \) in meters.
- The vertical axis represents acceleration \( a \) in meters per second squared \((\text{m/s}^2)\).
The acceleration starts from 2 \(\text{m/s}^2\) at \( s = 0\) and linearly increases to 4 \(\text{m/s}^2\) at \( s = 100 \text{ m}\).
**Multiple-Choice Options:**
1. 24.49 m/s
2. 15 m/s
3. 10 m/s
4. 25 m/s
**Explanation:**
The graph indicates a linear increase in acceleration from \( 2 \, \text{m/s}^2 \) to \( 4 \, \text{m/s}^2 \) over the distance of \( 100 \, \text{m} \).
To find the velocity at \( s = 100 \, \text{m} \), considering the initial velocity and the integration of acceleration, use the work-energy principle where the change in kinetic energy is equal to the work done by the net force.
The work done by the force is:
\[ W = \int F \, ds = \int ma \, ds \]
Since \( a \) is a function of \( s \):
\[ v_f^2 = v_0^2 + 2 \int a(s) \, ds \]
where \( v_f \) is the final velocity, \( v_0 \) is the initial velocity at \( s = 0 \), and \( a(s) \) is the acceleration as a function of \( s \).
Given the acceleration \( a(s) = 2 + \frac{2}{100}s\):
\[ \int_{0}^{100} a(s) \, ds = \int_{0}^{100} \left( 2 + \frac{2}{100}s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57026d2c-4d86-446e-bb74-abb5fa8f33cd%2F186b1c24-e695-4516-87e4-3b915180b3dc%2Fgyd7kr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Acceleration vs. Position of a Particle**
The question provided is:
*Acceleration vs. position of a particle is as given in the figure. If the velocity \( v = 5 \text{ m/s} \) at \( s = 0 \), what is the velocity at \( s = 100 \text{ m}?*
The diagram shows the acceleration \( a \) of a particle in relation to its position \( s \).
- The horizontal axis represents position \( s \) in meters.
- The vertical axis represents acceleration \( a \) in meters per second squared \((\text{m/s}^2)\).
The acceleration starts from 2 \(\text{m/s}^2\) at \( s = 0\) and linearly increases to 4 \(\text{m/s}^2\) at \( s = 100 \text{ m}\).
**Multiple-Choice Options:**
1. 24.49 m/s
2. 15 m/s
3. 10 m/s
4. 25 m/s
**Explanation:**
The graph indicates a linear increase in acceleration from \( 2 \, \text{m/s}^2 \) to \( 4 \, \text{m/s}^2 \) over the distance of \( 100 \, \text{m} \).
To find the velocity at \( s = 100 \, \text{m} \), considering the initial velocity and the integration of acceleration, use the work-energy principle where the change in kinetic energy is equal to the work done by the net force.
The work done by the force is:
\[ W = \int F \, ds = \int ma \, ds \]
Since \( a \) is a function of \( s \):
\[ v_f^2 = v_0^2 + 2 \int a(s) \, ds \]
where \( v_f \) is the final velocity, \( v_0 \) is the initial velocity at \( s = 0 \), and \( a(s) \) is the acceleration as a function of \( s \).
Given the acceleration \( a(s) = 2 + \frac{2}{100}s\):
\[ \int_{0}^{100} a(s) \, ds = \int_{0}^{100} \left( 2 + \frac{2}{100}s
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