Join the IDNLearn.com community and start finding the answers you need today. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
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.
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.
