Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Sure! Let's break down the problem step by step using the given information and the principles of linear motion at constant acceleration.
### Step 1: Understanding the Problem
We have a car that starts from rest (initial velocity = 0) and accelerates at [tex]\(2 \, \text{m/s}^2\)[/tex] for the first [tex]\(6 \, \text{seconds}\)[/tex]. Afterward, it travels at a constant velocity for half an hour. We need to find:
1. The distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex].
2. The distance traveled in the following half an hour.
### Step 2: Distance Traveled in the First 6 Seconds
First, we use the kinematic equation for distance ([tex]\(s\)[/tex]) under constant acceleration ([tex]\(a\)[/tex]):
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\(u\)[/tex] is the initial velocity,
- [tex]\(a\)[/tex] is the acceleration,
- [tex]\(t\)[/tex] is the time.
Given:
- [tex]\(u = 0 \, \text{m/s}\)[/tex] (since the car starts from rest),
- [tex]\(a = 2 \, \text{m/s}^2\)[/tex],
- [tex]\(t = 6 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ s = 0 \times 6 + \frac{1}{2} \times 2 \times (6)^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2} \times 2 \times 36 \][/tex]
[tex]\[ s = 36 \, \text{meters} \][/tex]
So, the distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(36 \, \text{meters}\)[/tex].
### Step 3: Velocity at the End of the First 6 Seconds
We need to find the velocity of the car at the end of [tex]\(6 \, \text{seconds}\)[/tex]. We use the kinematic equation for velocity ([tex]\(v\)[/tex]):
[tex]\[ v = u + at \][/tex]
Given:
- [tex]\(u = 0 \, \text{m/s}\)[/tex],
- [tex]\(a = 2 \, \text{m/s}^2\)[/tex],
- [tex]\(t = 6 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ v = 0 + 2 \times 6 \][/tex]
[tex]\[ v = 12 \, \text{m/s} \][/tex]
So, the velocity at the end of the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(12 \, \text{m/s}\)[/tex].
### Step 4: Distance Traveled in the Following Half Hour
Now, the car travels at a constant velocity of [tex]\(12 \, \text{m/s}\)[/tex] for half an hour. We need to convert this time into seconds for consistency.
[tex]\[ \text{Half an hour} = 0.5 \times 60 \times 60 \, \text{seconds} \][/tex]
[tex]\[ = 1800 \, \text{seconds} \][/tex]
The distance ([tex]\(d\)[/tex]) traveled at constant velocity is given by:
[tex]\[ d = vt \][/tex]
Given:
- [tex]\(v = 12 \, \text{m/s}\)[/tex],
- [tex]\(t = 1800 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ d = 12 \times 1800 \][/tex]
[tex]\[ d = 21600 \, \text{meters} \][/tex]
So, the distance traveled in the following half an hour is [tex]\(21600 \, \text{meters}\)[/tex].
### Summary
1. The distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(36 \, \text{meters}\)[/tex].
2. The distance traveled in the following half an hour is [tex]\(21600 \, \text{meters}\)[/tex].
These steps provide a detailed solution to the problem based on the principles of linear motion under constant acceleration and constant velocity.
### Step 1: Understanding the Problem
We have a car that starts from rest (initial velocity = 0) and accelerates at [tex]\(2 \, \text{m/s}^2\)[/tex] for the first [tex]\(6 \, \text{seconds}\)[/tex]. Afterward, it travels at a constant velocity for half an hour. We need to find:
1. The distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex].
2. The distance traveled in the following half an hour.
### Step 2: Distance Traveled in the First 6 Seconds
First, we use the kinematic equation for distance ([tex]\(s\)[/tex]) under constant acceleration ([tex]\(a\)[/tex]):
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\(u\)[/tex] is the initial velocity,
- [tex]\(a\)[/tex] is the acceleration,
- [tex]\(t\)[/tex] is the time.
Given:
- [tex]\(u = 0 \, \text{m/s}\)[/tex] (since the car starts from rest),
- [tex]\(a = 2 \, \text{m/s}^2\)[/tex],
- [tex]\(t = 6 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ s = 0 \times 6 + \frac{1}{2} \times 2 \times (6)^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2} \times 2 \times 36 \][/tex]
[tex]\[ s = 36 \, \text{meters} \][/tex]
So, the distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(36 \, \text{meters}\)[/tex].
### Step 3: Velocity at the End of the First 6 Seconds
We need to find the velocity of the car at the end of [tex]\(6 \, \text{seconds}\)[/tex]. We use the kinematic equation for velocity ([tex]\(v\)[/tex]):
[tex]\[ v = u + at \][/tex]
Given:
- [tex]\(u = 0 \, \text{m/s}\)[/tex],
- [tex]\(a = 2 \, \text{m/s}^2\)[/tex],
- [tex]\(t = 6 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ v = 0 + 2 \times 6 \][/tex]
[tex]\[ v = 12 \, \text{m/s} \][/tex]
So, the velocity at the end of the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(12 \, \text{m/s}\)[/tex].
### Step 4: Distance Traveled in the Following Half Hour
Now, the car travels at a constant velocity of [tex]\(12 \, \text{m/s}\)[/tex] for half an hour. We need to convert this time into seconds for consistency.
[tex]\[ \text{Half an hour} = 0.5 \times 60 \times 60 \, \text{seconds} \][/tex]
[tex]\[ = 1800 \, \text{seconds} \][/tex]
The distance ([tex]\(d\)[/tex]) traveled at constant velocity is given by:
[tex]\[ d = vt \][/tex]
Given:
- [tex]\(v = 12 \, \text{m/s}\)[/tex],
- [tex]\(t = 1800 \, \text{seconds}\)[/tex],
Substitute these values into the equation:
[tex]\[ d = 12 \times 1800 \][/tex]
[tex]\[ d = 21600 \, \text{meters} \][/tex]
So, the distance traveled in the following half an hour is [tex]\(21600 \, \text{meters}\)[/tex].
### Summary
1. The distance traveled in the first [tex]\(6 \, \text{seconds}\)[/tex] is [tex]\(36 \, \text{meters}\)[/tex].
2. The distance traveled in the following half an hour is [tex]\(21600 \, \text{meters}\)[/tex].
These steps provide a detailed solution to the problem based on the principles of linear motion under constant acceleration and constant velocity.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
