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Sagot :
Let's solve this problem step by step:
### 1. Distance Covered During the Initial Phase:
The car travels at a constant speed of 20 m/s for 10 seconds.
Distance = Speed × Time
[tex]\[ \text{Distance}_1 = 20 \, \text{m/s} \times 10 \, \text{s} = 200 \, \text{m} \][/tex]
### 2. Distance Covered During the Acceleration Phase:
The car accelerates at 3 m/s² for 6 seconds. We need to find the distance covered during this time.
First, we calculate the speed at the start of this phase, which is still 20 m/s.
Using the formula for distance during acceleration:
[tex]\[ \text{Distance} = ut + \frac{1}{2}at^2 \][/tex]
Where:
- \( u \) is the initial speed (20 m/s),
- \( a \) is the acceleration (3 m/s²),
- \( t \) is the time (6 seconds).
Plugging in the values:
[tex]\[ \text{Distance}_2 = 20 \, \text{m/s} \times 6 \, \text{s} + \frac{1}{2} \times 3 \, \text{m/s}^2 \times (6 \, \text{s})^2 \][/tex]
[tex]\[ \text{Distance}_2 = 120 \, \text{m} + 54 \, \text{m} = 174 \, \text{m} \][/tex]
Next, we calculate the speed at the end of the acceleration phase:
[tex]\[ \text{Final Speed} = \text{Initial Speed} + ( \text{Acceleration} \times \text{Time} ) \][/tex]
[tex]\[ \text{Final Speed} = 20 \, \text{m/s} + (3 \, \text{m/s}^2 \times 6 \, \text{s}) = 38 \, \text{m/s} \][/tex]
### 3. Time and Distance Covered During the Deceleration Phase:
The car decelerates at 4 m/s² until it stops.
First, we calculate the time taken to stop:
[tex]\[ \text{Final Speed} = \text{Initial Speed} + ( \text{Acceleration} \times \text{Time} ) \][/tex]
Since it decelerates to a stop (Final Speed = 0):
[tex]\[ 0 = 38 \, \text{m/s} + (-4 \, \text{m/s}^2 \times \text{Time}) \][/tex]
Solving for Time:
[tex]\[ \text{Time}_3 = \frac{38 \, \text{m/s}}{4 \, \text{m/s}^2} = 9.5 \, \text{s} \][/tex]
Now, we calculate the distance covered during deceleration:
[tex]\[ \text{Distance}_3 = \text{Initial Speed} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 \][/tex]
Since acceleration here is negative (deceleration), we substitute as follows:
[tex]\[ \text{Distance}_3 = 38 \, \text{m/s} \times 9.5 \, \text{s} + \frac{1}{2} \times (-4 \, \text{m/s}^2) \times (9.5 \, \text{s})^2 \][/tex]
[tex]\[ \text{Distance}_3 = 361 \, \text{m} - 180.5 \, \text{m} = 180.5 \, \text{m} \][/tex]
### 4. Total Distance and Time:
Now, we sum up all the distances and times.
Total Distance:
[tex]\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3 \][/tex]
[tex]\[ \text{Total Distance} = 200 \, \text{m} + 174 \, \text{m} + 180.5 \, \text{m} = 554.5 \, \text{m} \][/tex]
Total Time:
[tex]\[ \text{Total Time} = \text{Time}_1 + \text{Time}_2 + \text{Time}_3 \][/tex]
[tex]\[ \text{Total Time} = 10 \, \text{s} + 6 \, \text{s} + 9.5 \, \text{s} = 25.5 \, \text{s} \][/tex]
### Final Result:
- Total distance covered: 554.5 meters
- Total time taken: 25.5 seconds
### 1. Distance Covered During the Initial Phase:
The car travels at a constant speed of 20 m/s for 10 seconds.
Distance = Speed × Time
[tex]\[ \text{Distance}_1 = 20 \, \text{m/s} \times 10 \, \text{s} = 200 \, \text{m} \][/tex]
### 2. Distance Covered During the Acceleration Phase:
The car accelerates at 3 m/s² for 6 seconds. We need to find the distance covered during this time.
First, we calculate the speed at the start of this phase, which is still 20 m/s.
Using the formula for distance during acceleration:
[tex]\[ \text{Distance} = ut + \frac{1}{2}at^2 \][/tex]
Where:
- \( u \) is the initial speed (20 m/s),
- \( a \) is the acceleration (3 m/s²),
- \( t \) is the time (6 seconds).
Plugging in the values:
[tex]\[ \text{Distance}_2 = 20 \, \text{m/s} \times 6 \, \text{s} + \frac{1}{2} \times 3 \, \text{m/s}^2 \times (6 \, \text{s})^2 \][/tex]
[tex]\[ \text{Distance}_2 = 120 \, \text{m} + 54 \, \text{m} = 174 \, \text{m} \][/tex]
Next, we calculate the speed at the end of the acceleration phase:
[tex]\[ \text{Final Speed} = \text{Initial Speed} + ( \text{Acceleration} \times \text{Time} ) \][/tex]
[tex]\[ \text{Final Speed} = 20 \, \text{m/s} + (3 \, \text{m/s}^2 \times 6 \, \text{s}) = 38 \, \text{m/s} \][/tex]
### 3. Time and Distance Covered During the Deceleration Phase:
The car decelerates at 4 m/s² until it stops.
First, we calculate the time taken to stop:
[tex]\[ \text{Final Speed} = \text{Initial Speed} + ( \text{Acceleration} \times \text{Time} ) \][/tex]
Since it decelerates to a stop (Final Speed = 0):
[tex]\[ 0 = 38 \, \text{m/s} + (-4 \, \text{m/s}^2 \times \text{Time}) \][/tex]
Solving for Time:
[tex]\[ \text{Time}_3 = \frac{38 \, \text{m/s}}{4 \, \text{m/s}^2} = 9.5 \, \text{s} \][/tex]
Now, we calculate the distance covered during deceleration:
[tex]\[ \text{Distance}_3 = \text{Initial Speed} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 \][/tex]
Since acceleration here is negative (deceleration), we substitute as follows:
[tex]\[ \text{Distance}_3 = 38 \, \text{m/s} \times 9.5 \, \text{s} + \frac{1}{2} \times (-4 \, \text{m/s}^2) \times (9.5 \, \text{s})^2 \][/tex]
[tex]\[ \text{Distance}_3 = 361 \, \text{m} - 180.5 \, \text{m} = 180.5 \, \text{m} \][/tex]
### 4. Total Distance and Time:
Now, we sum up all the distances and times.
Total Distance:
[tex]\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3 \][/tex]
[tex]\[ \text{Total Distance} = 200 \, \text{m} + 174 \, \text{m} + 180.5 \, \text{m} = 554.5 \, \text{m} \][/tex]
Total Time:
[tex]\[ \text{Total Time} = \text{Time}_1 + \text{Time}_2 + \text{Time}_3 \][/tex]
[tex]\[ \text{Total Time} = 10 \, \text{s} + 6 \, \text{s} + 9.5 \, \text{s} = 25.5 \, \text{s} \][/tex]
### Final Result:
- Total distance covered: 554.5 meters
- Total time taken: 25.5 seconds
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