Q24. The acceleration of a particle undergoing rectilinear, erratic motion has acceleration-time (a-t) function shown in the image below. The equation for 0
Q24. The acceleration of a particle undergoing rectilinear, erratic motion has acceleration-time (a-t) function shown in the image below. The equation for 0
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
![**Question 24:**
The acceleration of a particle undergoing rectilinear, erratic motion is represented by the acceleration-time (\(a-t\)) function shown in the image below. For \(0 < t < 3\) seconds (the curved portion), the equation is \(a = C \cdot t^2 \, \text{m/s}^2\), where \(C\) is a constant, and \(t\) is the time in seconds. If the constant \(C\) is 0.1, and the initial velocity \(v_0 = 0\) at time \(t = 0\), determine the particle's velocity when \(t = 5.4\) seconds. Please pay attention: the numbers may change since they are randomized. Your answer must include three decimal places and proper SI unit.
**Graph Explanation:**
The graph depicts the acceleration (\(a\) in m/s²) versus time (\(t\) in seconds). It shows:
- A rising curve from \(t = 0\) to \(t = 3\) seconds, where acceleration is increasing.
- Beyond \(t = 3\) seconds, the graph shows a downward-sloping line till \(t = 6\) seconds, indicating decreasing acceleration.
The graph implies that for \(0 < t < 3\) seconds, the acceleration follows a quadratic relation \(a = C \cdot t^2\).
**Your Answer:**
| Answer | Units |
|-------------|-------|
| [Numeric Answer] | [m/s] |](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff360a777-96c9-45b8-b8a4-fa76e1e19cb5%2Fda58d2e3-735a-4390-ae3b-b376770cc4c0%2Fbfell7j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 24:**
The acceleration of a particle undergoing rectilinear, erratic motion is represented by the acceleration-time (\(a-t\)) function shown in the image below. For \(0 < t < 3\) seconds (the curved portion), the equation is \(a = C \cdot t^2 \, \text{m/s}^2\), where \(C\) is a constant, and \(t\) is the time in seconds. If the constant \(C\) is 0.1, and the initial velocity \(v_0 = 0\) at time \(t = 0\), determine the particle's velocity when \(t = 5.4\) seconds. Please pay attention: the numbers may change since they are randomized. Your answer must include three decimal places and proper SI unit.
**Graph Explanation:**
The graph depicts the acceleration (\(a\) in m/s²) versus time (\(t\) in seconds). It shows:
- A rising curve from \(t = 0\) to \(t = 3\) seconds, where acceleration is increasing.
- Beyond \(t = 3\) seconds, the graph shows a downward-sloping line till \(t = 6\) seconds, indicating decreasing acceleration.
The graph implies that for \(0 < t < 3\) seconds, the acceleration follows a quadratic relation \(a = C \cdot t^2\).
**Your Answer:**
| Answer | Units |
|-------------|-------|
| [Numeric Answer] | [m/s] |
Expert Solution

Step 1
Given data
For 0 to 3 sec
Acceleration = C.t2
Where C= 0.1
Initial velocity= 0
Find the velocity at t= 5.4 sec
Step by step
Solved in 2 steps with 1 images

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