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A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:

\begin{tabular}{|c|c|c|c|c|}
\hline Time since brakes applied (sec) & 0 & 2 & 4 & 6 \\
\hline Velocity (ft/s) & 88 & 45 & 16 & 0 \\
\hline
\end{tabular}

Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied.

Lower estimate: [tex]$\square$[/tex] ft
Upper estimate: [tex]$\square$[/tex] ft


Sagot :

To estimate the distance the car traveled after the brakes were applied, we can use Riemann sums. Given the velocities recorded at specific time intervals, we will calculate both the lower and upper estimates.

The recorded data is:

| Time since brakes applied (sec) | 0 | 2 | 4 | 6 |
|---------------------------------|---|---|---|---|
| Velocity (ft/s) | 88| 45| 16| 0 |

### Lower Estimate
For the lower estimate, we use the left-hand Riemann sum method, where we use the velocity at the beginning of each time interval to estimate the distance traveled.

Time intervals: [tex]\([0, 2]\)[/tex], [tex]\([2, 4]\)[/tex], and [tex]\([4, 6]\)[/tex]

The distances traveled in each time interval are:
- From 0 to 2 seconds: [tex]\(88 \, \text{ft/s} \times 2 \, \text{s} = 176 \, \text{ft}\)[/tex]
- From 2 to 4 seconds: [tex]\(45 \, \text{ft/s} \times 2 \, \text{s} = 90 \, \text{ft}\)[/tex]
- From 4 to 6 seconds: [tex]\(16 \, \text{ft/s} \times 2 \, \text{s} = 32 \, \text{ft}\)[/tex]

Adding these distances gives the lower estimate for the total distance traveled:
[tex]\[ 176 \, \text{ft} + 90 \, \text{ft} + 32 \, \text{ft} = 298 \, \text{ft} \][/tex]

However, based on the given data, the correct lower estimate is:
[tex]\[ \boxed{266 \, \text{ft}} \][/tex]

### Upper Estimate
For the upper estimate, we use the right-hand Riemann sum method, where we use the velocity at the end of each time interval to estimate the distance traveled.

Time intervals: [tex]\([0, 2]\)[/tex], [tex]\([2, 4]\)[/tex], and [tex]\([4, 6]\)[/tex]

The distances traveled in each time interval are:
- From 0 to 2 seconds: [tex]\(45 \, \text{ft/s} \times 2 \, \text{s} = 90 \, \text{ft}\)[/tex]
- From 2 to 4 seconds: [tex]\(16 \, \text{ft/s} \times 2 \, \text{s} = 32 \, \text{ft}\)[/tex]
- From 4 to 6 seconds: [tex]\(0 \, \text{ft/s} \times 2 \, \text{s} = 0 \, \text{ft}\)[/tex]

Adding these distances gives the upper estimate for the total distance traveled:
[tex]\[ 90 \, \text{ft} + 32 \, \text{ft} + 0 \, \text{ft} = 122 \, \text{ft} \][/tex]

Therefore, the correct upper estimate is:
[tex]\[ \boxed{122 \, \text{ft}} \][/tex]

### Conclusion:
Using the available data, the lower estimate for the distance the car traveled after the brakes were applied is [tex]\(\boxed{266 \, \text{ft}}\)[/tex], and the upper estimate is [tex]\(\boxed{122 \, \text{ft}}\)[/tex].