Question: Racing cars driven by Chris and Kelly are side by side at the start of a race. The table shows the velocities of each car (in miles per hour) during the first ten seconds of the race.
t v_C v_K t v_C v_K
0 0 0 6 68 79
1 21 23 7 75 86
2 33 36 8 81 93
3 47 52 9 86 98
4 54 60 10 91 103
5 63 72
Use the midpoint rule to estimate how much farther (in mi) Kelly travels than Chris does during the first ten seconds. (Round your answer to four decimal places.)
1. **State the problem:** We want to estimate how much farther Kelly travels than Chris during the first 10 seconds using the midpoint rule on their velocity data.
2. **Recall the midpoint rule formula:**
$$\text{Distance} \approx \Delta t \sum_{i=1}^n v\left(t_{i-\frac{1}{2}}\right)$$
where $\Delta t$ is the time interval and $v\left(t_{i-\frac{1}{2}}\right)$ is the velocity at the midpoint of each subinterval.
3. **Given data:** Velocities $v_C$ and $v_K$ at times $t=0,1,2,...,10$ seconds.
4. **Calculate midpoints and their velocities:**
Midpoints are at $t=0.5,1.5,2.5,...,9.5$ seconds.
Midpoint velocities for Chris $v_C$:
- $v_C(0.5) = \frac{v_C(0)+v_C(1)}{2} = \frac{0+21}{2} = 10.5$
- $v_C(1.5) = \frac{21+33}{2} = 27$
- $v_C(2.5) = \frac{33+47}{2} = 40$
- $v_C(3.5) = \frac{47+54}{2} = 50.5$
- $v_C(4.5) = \frac{54+63}{2} = 58.5$
- $v_C(5.5) = \frac{63+68}{2} = 65.5$
- $v_C(6.5) = \frac{68+75}{2} = 71.5$
- $v_C(7.5) = \frac{75+81}{2} = 78$
- $v_C(8.5) = \frac{81+86}{2} = 83.5$
- $v_C(9.5) = \frac{86+91}{2} = 88.5$
Midpoint velocities for Kelly $v_K$:
- $v_K(0.5) = \frac{0+23}{2} = 11.5$
- $v_K(1.5) = \frac{23+36}{2} = 29.5$
- $v_K(2.5) = \frac{36+52}{2} = 44$
- $v_K(3.5) = \frac{52+60}{2} = 56$
- $v_K(4.5) = \frac{60+72}{2} = 66$
- $v_K(5.5) = \frac{72+79}{2} = 75.5$
- $v_K(6.5) = \frac{79+86}{2} = 82.5$
- $v_K(7.5) = \frac{86+93}{2} = 89.5$
- $v_K(8.5) = \frac{93+98}{2} = 95.5$
- $v_K(9.5) = \frac{98+103}{2} = 100.5$
5. **Calculate distance traveled by each car:**
Time interval $\Delta t = 1$ second.
Distance Chris:
$$\sum v_C(\text{midpoints}) = 10.5 + 27 + 40 + 50.5 + 58.5 + 65.5 + 71.5 + 78 + 83.5 + 88.5 = 573.5$$
Distance Kelly:
$$\sum v_K(\text{midpoints}) = 11.5 + 29.5 + 44 + 56 + 66 + 75.5 + 82.5 + 89.5 + 95.5 + 100.5 = 650.5$$
6. **Convert velocities from miles per hour to miles per second:**
Since velocities are in miles per hour, and time is in seconds, convert by dividing by 3600:
$$v_{\text{miles/sec}} = \frac{v_{\text{miles/hour}}}{3600}$$
7. **Calculate distances in miles:**
$$\text{Distance}_C = \frac{573.5}{3600} \times 1 = 0.1593 \text{ miles}$$
$$\text{Distance}_K = \frac{650.5}{3600} \times 1 = 0.1807 \text{ miles}$$
8. **Find how much farther Kelly travels than Chris:**
$$0.1807 - 0.1593 = 0.0214 \text{ miles}$$
9. **Round to four decimal places:**
$$\boxed{0.0214}$$ miles farther.
**Note:** The user-provided answer was 0.0231, but following the midpoint rule and given data, the calculated difference is approximately 0.0214 miles.