Question

conclusion

1. if you were driving a car and suddenly saw a child run out into the road...how far would your car travel before you react and start to break. if you are going 45mph (20.1168 m/s).

a. use your reaction time without distractions:
( d = v_i t )
b. use your reaction time with distractions:
chart with columns: average reaction time (without distractions) from table 1 (s), average reaction time (with distractions) from table 2 (s), difference in times; rows with numbers (e.g., 0.41, 0.42, 0.01; 0.36, 0.40, 0.04; etc.)

Explanation:

Step1: Identify the formula and values

We use the formula \( d = v_i t \), where \( v_i = 20.1168 \, \text{m/s} \) (initial velocity) and we need the average reaction time without distractions. From the table, the average reaction time without distractions (from Table 1) – let's take the average of the given times (0.41, 0.36, 0.38, 0.35). First, calculate the average: \( \frac{0.41 + 0.36 + 0.38 + 0.35}{4} = \frac{1.5}{4} = 0.375 \, \text{s} \).

Step2: Substitute into the formula

Now substitute \( v_i = 20.1168 \, \text{m/s} \) and \( t = 0.375 \, \text{s} \) into \( d = v_i t \).
\( d = 20.1168 \times 0.375 \)
Calculate the product: \( 20.1168 \times 0.375 = 7.5438 \, \text{m} \) (approximate, depending on the exact average reaction time used; if we use one of the individual times, say the first average reaction time without distractions as 0.41s: \( d = 20.1168 \times 0.41 \approx 8.2479 \, \text{m} \); but since the table has multiple values, the average is a better approach. Let's check the table again – maybe the "Average reaction time from Table 1 (without distractions)" has an average. Wait, the table shows:

First row: Average reaction time (without distractions) from Table 1: 0.41s

Second: 0.36s

Third: 0.38s

Fourth: 0.35s

So average is \( (0.41 + 0.36 + 0.38 + 0.35)/4 = 1.5/4 = 0.375 \, \text{s} \).

But maybe the problem expects using the average reaction time without distractions as, say, the first value? Wait, the problem says "use your reaction time without distractions" – maybe the average of the "Average reaction time (without distractions)" column. Let's proceed with the formula.

If we take \( t = 0.375 \, \text{s} \) (average) and \( v_i = 20.1168 \, \text{m/s} \), then \( d = 20.1168 \times 0.375 = 7.5438 \, \text{m} \).

Alternatively, if we take one of the individual times, like 0.41s (first row): \( d = 20.1168 \times 0.41 \approx 8.248 \, \text{m} \).

But let's check the table structure again. The table has columns:

• Average reaction time (without distractions) from Table 1 (s): 0.41, 0.36, 0.38, 0.35

• Average reaction time (with distractions) from Table 2 (s): 0.42, 0.40, 0.41, 0.34

• Difference in times: 0.01, 0.04, 0.03, 0.01

So for part a, "use your reaction time without distractions" – probably the average of the "without distractions" column. So average \( t = (0.41 + 0.36 + 0.38 + 0.35)/4 = 1.5/4 = 0.375 \, \text{s} \).

Then \( d = v_i t = 20.1168 \, \text{m/s} \times 0.375 \, \text{s} = 7.5438 \, \text{m} \approx 7.54 \, \text{m} \) (or more precisely, 7.5438 m).

Answer:

Approximately \( \boldsymbol{7.54 \, \text{m}} \) (or depending on the reaction time used, e.g., if using 0.41s, it's approximately 8.25 m).