Answer :
Certainly! Let's go through each part of the question in detail.
### Part 6a:
#### i) Uniform Motion:
Uniform motion refers to motion at a constant velocity, which means there is no acceleration. The graph of velocity versus time for uniform motion is a horizontal straight line.
Sketch:
```
v (m/s)
| _______
| |
| |
| |
|______|____________________ t (s)
```
#### ii) Uniform Acceleration:
Uniform acceleration refers to motion where the velocity increases linearly over time. This implies a constant rate of change of velocity, and the graph of velocity versus time is a straight line with a positive slope.
Sketch:
```
v (m/s)
|
|
|
|
| /
| /
|/__________________________ t (s)
```
### Part 6b:
The table provided shows the velocity of a car at different times:
```
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline Timels & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 & 12.0 \\
\hline V/mos -1 & 25.0 & 45.0 & 65.0 & 75.0 & 75.0 & 75.0 \\
\hline
\end{tabular}
```
#### Part i) Plot the velocity-time graph:
To plot the velocity-time graph, you can use the points provided:
Velocity-Time Graph:
```
v (m/s)
| x
| |
| x x |
| x |
| x |
| x x |
|_____________________________ t (s)
2 4 6 8 10 12
```
Here, each "x" marks the corresponding points: (2, 25), (4, 45), (6, 65), (8, 75), (10, 75), (12, 75). Connect these points with lines to visualize the motion.
#### Part ii) Determine specific values from the graph:
(a) Initial velocity of the car:
The initial velocity is the car's velocity at the start of the time interval given in the table, which is at [tex]\( t = 2.0 \)[/tex] seconds.
From the table,
[tex]\[ \text{Initial velocity} = 25.0 \, \text{m/s} \][/tex]
(b) Acceleration of the car:
Acceleration is defined as the rate of change of velocity with respect to time. We can calculate it using the following formula:
[tex]\[ a = \frac{\Delta v}{\Delta t} \][/tex]
We will consider the acceleration between [tex]\( t = 2.0 \)[/tex] seconds and [tex]\( t = 6.0 \)[/tex] seconds:
[tex]\[ \Delta v = v(6.0 \, s) - v(2.0 \, s) = 65.0 \, \text{m/s} - 25.0 \, \text{m/s} = 40.0 \, \text{m/s} \][/tex]
[tex]\[ \Delta t = 6.0 \, \text{s} - 2.0 \, \text{s} = 4.0 \, \text{s} \][/tex]
[tex]\[ a = \frac{40.0 \, \text{m/s}}{4.0 \, \text{s}} = 10.0 \, \text{m/s}^2 \][/tex]
(c) Distance covered by the car in 9 seconds:
To find the distance covered in 9 seconds, we will use the area under the velocity-time graph up to [tex]\( t = 9.0 \)[/tex] seconds.
- From [tex]\( t = 2 \, \text{s} \)[/tex] to [tex]\( t = 6 \, \text{s} \)[/tex], the graph is a trapezoid.
- From [tex]\( t = 6 \, \text{s} \)[/tex] to [tex]\( t = 8 \, \text{s} \)[/tex], the graph is a trapezoid.
- From [tex]\( t = 8 \, \text{s} \)[/tex] to [tex]\( t = 9 \, \text{s} \)[/tex], the graph is a rectangle.
Let's break down the calculation:
1. Area from [tex]\( t = 2 \, \text{s} \)[/tex] to [tex]\( t = 6 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \times (v(2 \, \text{s}) + v(6 \, \text{s})) \times (\Delta t) \][/tex]
[tex]\[ = \frac{1}{2} \times (25.0 \, \text{m/s} + 65.0 \, \text{m/s}) \times 4.0 \, \text{s} = \frac{1}{2} \times 90.0 \, \text{m/s} \times 4.0 \, \text{s} = 180.0 \, \text{m} \][/tex]
2. Area from [tex]\( t = 6 \, \text{s} \)[/tex] to [tex]\( t = 8 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \times (v(6 \, \text{s}) + v(8 \, \text{s})) \times (\Delta t) \][/tex]
[tex]\[ = \frac{1}{2} \times (65.0 \, \text{m/s} + 75.0 \, \text{m/s}) \times 2.0 \, \text{s} = \frac{1}{2} \times 140.0 \, \text{m/s} \times 2.0 \, \text{s} = 140.0 \, \text{m} \][/tex]
3. Area from [tex]\( t = 8 \, \text{s} \)[/tex] to [tex]\( t = 9 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = v(8 \, \text{s}) \times (\Delta t) \][/tex]
[tex]\[ = 75.0 \, \text{m/s} \times 1.0 \, \text{s} = 75.0 \, \text{m} \][/tex]
Total Distance covered in 9 seconds:
[tex]\[ \text{Total Distance} = 180.0 \, \text{m} + 140.0 \, \text{m} + 75.0 \, \text{m} = 395.0 \, \text{m} \][/tex]
Thus, the distance covered by the car in the first 9 seconds is [tex]\( 395.0 \, \text{m} \)[/tex].
### Part 6a:
#### i) Uniform Motion:
Uniform motion refers to motion at a constant velocity, which means there is no acceleration. The graph of velocity versus time for uniform motion is a horizontal straight line.
Sketch:
```
v (m/s)
| _______
| |
| |
| |
|______|____________________ t (s)
```
#### ii) Uniform Acceleration:
Uniform acceleration refers to motion where the velocity increases linearly over time. This implies a constant rate of change of velocity, and the graph of velocity versus time is a straight line with a positive slope.
Sketch:
```
v (m/s)
|
|
|
|
| /
| /
|/__________________________ t (s)
```
### Part 6b:
The table provided shows the velocity of a car at different times:
```
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline Timels & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 & 12.0 \\
\hline V/mos -1 & 25.0 & 45.0 & 65.0 & 75.0 & 75.0 & 75.0 \\
\hline
\end{tabular}
```
#### Part i) Plot the velocity-time graph:
To plot the velocity-time graph, you can use the points provided:
Velocity-Time Graph:
```
v (m/s)
| x
| |
| x x |
| x |
| x |
| x x |
|_____________________________ t (s)
2 4 6 8 10 12
```
Here, each "x" marks the corresponding points: (2, 25), (4, 45), (6, 65), (8, 75), (10, 75), (12, 75). Connect these points with lines to visualize the motion.
#### Part ii) Determine specific values from the graph:
(a) Initial velocity of the car:
The initial velocity is the car's velocity at the start of the time interval given in the table, which is at [tex]\( t = 2.0 \)[/tex] seconds.
From the table,
[tex]\[ \text{Initial velocity} = 25.0 \, \text{m/s} \][/tex]
(b) Acceleration of the car:
Acceleration is defined as the rate of change of velocity with respect to time. We can calculate it using the following formula:
[tex]\[ a = \frac{\Delta v}{\Delta t} \][/tex]
We will consider the acceleration between [tex]\( t = 2.0 \)[/tex] seconds and [tex]\( t = 6.0 \)[/tex] seconds:
[tex]\[ \Delta v = v(6.0 \, s) - v(2.0 \, s) = 65.0 \, \text{m/s} - 25.0 \, \text{m/s} = 40.0 \, \text{m/s} \][/tex]
[tex]\[ \Delta t = 6.0 \, \text{s} - 2.0 \, \text{s} = 4.0 \, \text{s} \][/tex]
[tex]\[ a = \frac{40.0 \, \text{m/s}}{4.0 \, \text{s}} = 10.0 \, \text{m/s}^2 \][/tex]
(c) Distance covered by the car in 9 seconds:
To find the distance covered in 9 seconds, we will use the area under the velocity-time graph up to [tex]\( t = 9.0 \)[/tex] seconds.
- From [tex]\( t = 2 \, \text{s} \)[/tex] to [tex]\( t = 6 \, \text{s} \)[/tex], the graph is a trapezoid.
- From [tex]\( t = 6 \, \text{s} \)[/tex] to [tex]\( t = 8 \, \text{s} \)[/tex], the graph is a trapezoid.
- From [tex]\( t = 8 \, \text{s} \)[/tex] to [tex]\( t = 9 \, \text{s} \)[/tex], the graph is a rectangle.
Let's break down the calculation:
1. Area from [tex]\( t = 2 \, \text{s} \)[/tex] to [tex]\( t = 6 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \times (v(2 \, \text{s}) + v(6 \, \text{s})) \times (\Delta t) \][/tex]
[tex]\[ = \frac{1}{2} \times (25.0 \, \text{m/s} + 65.0 \, \text{m/s}) \times 4.0 \, \text{s} = \frac{1}{2} \times 90.0 \, \text{m/s} \times 4.0 \, \text{s} = 180.0 \, \text{m} \][/tex]
2. Area from [tex]\( t = 6 \, \text{s} \)[/tex] to [tex]\( t = 8 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \times (v(6 \, \text{s}) + v(8 \, \text{s})) \times (\Delta t) \][/tex]
[tex]\[ = \frac{1}{2} \times (65.0 \, \text{m/s} + 75.0 \, \text{m/s}) \times 2.0 \, \text{s} = \frac{1}{2} \times 140.0 \, \text{m/s} \times 2.0 \, \text{s} = 140.0 \, \text{m} \][/tex]
3. Area from [tex]\( t = 8 \, \text{s} \)[/tex] to [tex]\( t = 9 \, \text{s} \)[/tex]:
[tex]\[ \text{Area} = v(8 \, \text{s}) \times (\Delta t) \][/tex]
[tex]\[ = 75.0 \, \text{m/s} \times 1.0 \, \text{s} = 75.0 \, \text{m} \][/tex]
Total Distance covered in 9 seconds:
[tex]\[ \text{Total Distance} = 180.0 \, \text{m} + 140.0 \, \text{m} + 75.0 \, \text{m} = 395.0 \, \text{m} \][/tex]
Thus, the distance covered by the car in the first 9 seconds is [tex]\( 395.0 \, \text{m} \)[/tex].
