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Sagot :
To solve this problem, we'll use the information provided to estimate the distance the car traveled after the brakes were applied. We can use the left Riemann sum and the right Riemann sum to find lower and upper estimates for the distance traveled.
### Given Data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (sec)} & \text{Velocity (ft/sec)} \\ \hline 0 & 88 \\ 2 & 45 \\ 4 & 16 \\ 6 & 0 \\ \hline \end{array} \][/tex]
### Explanation:
We need to estimate the distance traveled by the car in the interval [0, 6] seconds. Each time interval and corresponding velocity can be used for Riemann sums.
#### Lower Estimate:
For the lower estimate, we'll use the left Riemann sum, which means using the velocity at the start of each interval.
1. Interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 2 \)[/tex]:
[tex]\[ \text{Velocity} = 88 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 88 \times 2 = 176 \, \text{ft} \][/tex]
2. Interval from [tex]\( t = 2 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Velocity} = 45 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 45 \times 2 = 90 \, \text{ft} \][/tex]
3. Interval from [tex]\( t = 4 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \text{Velocity} = 16 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 16 \times 2 = 32 \, \text{ft} \][/tex]
Adding these distances gives us the lower estimate:
[tex]\[ \text{Lower Estimate} = 176 + 90 + 32 = 298 \, \text{ft} \][/tex]
#### Upper Estimate:
For the upper estimate, we'll use the right Riemann sum, which means using the velocity at the end of each interval.
1. Interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 2 \)[/tex]:
[tex]\[ \text{Velocity} = 45 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 45 \times 2 = 90 \, \text{ft} \][/tex]
2. Interval from [tex]\( t = 2 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Velocity} = 16 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 16 \times 2 = 32 \, \text{ft} \][/tex]
3. Interval from [tex]\( t = 4 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \text{Velocity} = 0 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 0 \times 2 = 0 \, \text{ft} \][/tex]
Adding these distances gives us the upper estimate:
[tex]\[ \text{Upper Estimate} = 90 + 32 + 0 = 122 \, \text{ft} \][/tex]
### Summary:
[tex]\[ \text{Lower Estimate} = 298 \, \text{ft} \][/tex]
[tex]\[ \text{Upper Estimate} = 122 \, \text{ft} \][/tex]
### Given Data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (sec)} & \text{Velocity (ft/sec)} \\ \hline 0 & 88 \\ 2 & 45 \\ 4 & 16 \\ 6 & 0 \\ \hline \end{array} \][/tex]
### Explanation:
We need to estimate the distance traveled by the car in the interval [0, 6] seconds. Each time interval and corresponding velocity can be used for Riemann sums.
#### Lower Estimate:
For the lower estimate, we'll use the left Riemann sum, which means using the velocity at the start of each interval.
1. Interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 2 \)[/tex]:
[tex]\[ \text{Velocity} = 88 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 88 \times 2 = 176 \, \text{ft} \][/tex]
2. Interval from [tex]\( t = 2 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Velocity} = 45 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 45 \times 2 = 90 \, \text{ft} \][/tex]
3. Interval from [tex]\( t = 4 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \text{Velocity} = 16 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 16 \times 2 = 32 \, \text{ft} \][/tex]
Adding these distances gives us the lower estimate:
[tex]\[ \text{Lower Estimate} = 176 + 90 + 32 = 298 \, \text{ft} \][/tex]
#### Upper Estimate:
For the upper estimate, we'll use the right Riemann sum, which means using the velocity at the end of each interval.
1. Interval from [tex]\( t = 0 \)[/tex] to [tex]\( t = 2 \)[/tex]:
[tex]\[ \text{Velocity} = 45 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 45 \times 2 = 90 \, \text{ft} \][/tex]
2. Interval from [tex]\( t = 2 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Velocity} = 16 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 16 \times 2 = 32 \, \text{ft} \][/tex]
3. Interval from [tex]\( t = 4 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \text{Velocity} = 0 \, \text{ft/sec} \quad \text{and} \quad \Delta t = 2 \, \text{sec} \][/tex]
[tex]\[ \text{Distance} = 0 \times 2 = 0 \, \text{ft} \][/tex]
Adding these distances gives us the upper estimate:
[tex]\[ \text{Upper Estimate} = 90 + 32 + 0 = 122 \, \text{ft} \][/tex]
### Summary:
[tex]\[ \text{Lower Estimate} = 298 \, \text{ft} \][/tex]
[tex]\[ \text{Upper Estimate} = 122 \, \text{ft} \][/tex]
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