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Sagot :
To address the problem of the car's motion, let's break it down step by step and understand how the car's velocity changes over time.
1. Initial Deceleration:
- The car starts at an initial velocity of [tex]\( 20 \, \text{m/s} \)[/tex].
- It decelerates uniformly to rest ([tex]\(0 \, \text{m/s}\)[/tex]) over 4 seconds.
- The acceleration during this period can be calculated as:
[tex]\[ a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} = \frac{0 \, \text{m/s} - 20 \, \text{m/s}}{4 \, \text{s}} = -5 \, \text{m/s}^2 \][/tex]
2. Uniform Reverse Acceleration:
- After coming to rest, the car reverses with uniform acceleration.
- It reaches the same speed but in the opposite direction ([tex]\(-20 \, \text{m/s}\)[/tex]) in another 4 seconds.
- The acceleration during this period is:
[tex]\[ a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} = \frac{-20 \, \text{m/s} - 0 \, \text{m/s}}{4 \, \text{s}} = -5 \, \text{m/s}^2 \][/tex]
From this information, we can outline the key points to sketch the velocity-time graph:
- At [tex]\( t = 0 \)[/tex] seconds, the velocity is [tex]\( 20 \, \text{m/s} \)[/tex].
- At [tex]\( t = 4 \)[/tex] seconds, the velocity is [tex]\( 0 \, \text{m/s} \)[/tex] (as it has stopped).
- At [tex]\( t = 8 \)[/tex] seconds, the velocity is [tex]\( -20 \, \text{m/s} \)[/tex] (reached its maximum reverse speed).
### Sketching the Velocity-Time Graph:
- Time Points: [tex]\(0\)[/tex], [tex]\(4\)[/tex], and [tex]\(8\)[/tex] seconds.
- Velocity Points: [tex]\(20 \, \text{m/s}\)[/tex], [tex]\(0 \, \text{m/s}\)[/tex], and [tex]\(-20 \, \text{m/s}\)[/tex].
### Graph:
1. From [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex] seconds:
- The line should slope downwards linearly from [tex]\( 20 \, \text{m/s} \)[/tex] to [tex]\( 0 \, \text{m/s} \)[/tex].
2. From [tex]\( t = 4 \)[/tex] to [tex]\( t = 8 \)[/tex] seconds:
- The line should continue sloping downwards linearly from [tex]\( 0 \, \text{m/s} \)[/tex] to [tex]\( -20 \, \text{m/s} \)[/tex].
Here's how you can visualize it:
[tex]\[ \begin{array}{c|c} \text{Time (s)} & \text{Velocity (m/s)} \\ \hline 0 & 20 \\ 4 & 0 \\ 8 & -20 \\ \end{array} \][/tex]
Plot these points on a graph with time on the x-axis and velocity on the y-axis, then connect them with straight lines to show the uniform deceleration and acceleration.
The graph should look like this:
```
Velocity (m/s)
20 |
| /
| /
| /
| /
0 |--------------------------------- Time (s)
| \
| \
| \
| \
-20 |
+-----------------------------------
0 4 8
```
This sketch clearly illustrates the uniform deceleration to rest and the uniform acceleration in the reverse direction.
1. Initial Deceleration:
- The car starts at an initial velocity of [tex]\( 20 \, \text{m/s} \)[/tex].
- It decelerates uniformly to rest ([tex]\(0 \, \text{m/s}\)[/tex]) over 4 seconds.
- The acceleration during this period can be calculated as:
[tex]\[ a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} = \frac{0 \, \text{m/s} - 20 \, \text{m/s}}{4 \, \text{s}} = -5 \, \text{m/s}^2 \][/tex]
2. Uniform Reverse Acceleration:
- After coming to rest, the car reverses with uniform acceleration.
- It reaches the same speed but in the opposite direction ([tex]\(-20 \, \text{m/s}\)[/tex]) in another 4 seconds.
- The acceleration during this period is:
[tex]\[ a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} = \frac{-20 \, \text{m/s} - 0 \, \text{m/s}}{4 \, \text{s}} = -5 \, \text{m/s}^2 \][/tex]
From this information, we can outline the key points to sketch the velocity-time graph:
- At [tex]\( t = 0 \)[/tex] seconds, the velocity is [tex]\( 20 \, \text{m/s} \)[/tex].
- At [tex]\( t = 4 \)[/tex] seconds, the velocity is [tex]\( 0 \, \text{m/s} \)[/tex] (as it has stopped).
- At [tex]\( t = 8 \)[/tex] seconds, the velocity is [tex]\( -20 \, \text{m/s} \)[/tex] (reached its maximum reverse speed).
### Sketching the Velocity-Time Graph:
- Time Points: [tex]\(0\)[/tex], [tex]\(4\)[/tex], and [tex]\(8\)[/tex] seconds.
- Velocity Points: [tex]\(20 \, \text{m/s}\)[/tex], [tex]\(0 \, \text{m/s}\)[/tex], and [tex]\(-20 \, \text{m/s}\)[/tex].
### Graph:
1. From [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex] seconds:
- The line should slope downwards linearly from [tex]\( 20 \, \text{m/s} \)[/tex] to [tex]\( 0 \, \text{m/s} \)[/tex].
2. From [tex]\( t = 4 \)[/tex] to [tex]\( t = 8 \)[/tex] seconds:
- The line should continue sloping downwards linearly from [tex]\( 0 \, \text{m/s} \)[/tex] to [tex]\( -20 \, \text{m/s} \)[/tex].
Here's how you can visualize it:
[tex]\[ \begin{array}{c|c} \text{Time (s)} & \text{Velocity (m/s)} \\ \hline 0 & 20 \\ 4 & 0 \\ 8 & -20 \\ \end{array} \][/tex]
Plot these points on a graph with time on the x-axis and velocity on the y-axis, then connect them with straight lines to show the uniform deceleration and acceleration.
The graph should look like this:
```
Velocity (m/s)
20 |
| /
| /
| /
| /
0 |--------------------------------- Time (s)
| \
| \
| \
| \
-20 |
+-----------------------------------
0 4 8
```
This sketch clearly illustrates the uniform deceleration to rest and the uniform acceleration in the reverse direction.
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