Question

passage 1
students used 2 methods to calculate d, a cars total stopping distance; d is the distance a car travels from the time a driver first reacts to an emergency until the car comes to a complete stop.
in method 1, r is the distance a car travels during a drivers assumed reaction time of 0.75 sec, and b is the average distance traveled once the brakes are applied. method 2 assumes that d = initial speed in ft/sec * 2 sec. table 1 lists r, b, and d for various initial speeds, where d was computed using both methods. figure 1 contains graphs of d versus initial speed for method 1 and method 2.
in method 1, d equals
f. r + b
g. r - b
h. r × b
j. r + b

Explanation:

Step 1: Analyze the given data for Method 1

We have the values of \( R \), \( B \), and \( D \) for different initial speeds. Let's take the first row: \( R = 22 \) ft, \( B = 20 \) ft, \( D = 42 \) ft. Let's check the operations:

• Option F: \( R + B=22 + 20 = 42\), which matches \( D \).
• Option G: \( R - B=22-20 = 2

eq42\)

• Option H: \( R\times B = 22\times20=440

eq42\)

• Option J: \( R\div B=\frac{22}{20} = 1.1

eq42\) (Wait, the option J is \( R\div B \)? Wait, the original options: F. \( R + B \), G. \( R - B \), H. \( R\times B \), J. \( R + B \)? Wait, maybe a typo, but in the first row, \( R + B = 22+20 = 42=D \). Let's check another row. For initial speed 40 mi/hr: \( R = 44 \), \( B = 80 \), \( D = 124 \). \( 44+80=124 \), which matches. Third row: \( R = 66 \), \( B = 180 \), \( D = 246 \). \( 66 + 180=246 \). Fourth row: \( R = 88 \), \( B = 320 \), \( D = 408 \). \( 88+320 = 408 \). So the operation is \( R + B \).

Answer:

F. \( R + B \)