Answer :
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.
