Question

The expressions in the table above give the distance of each of two trains from Centerville at t hours after 12:00 noon. At what time will the trains be equidistant from Centerville?
Freight Train: \(-10t + 115\)
Passenger Train: \(-20t + 150\)

• A. 1:30 p.m.
• B. 3:30 p.m.
• C. 5:10 p.m.
• D. 8:50 p.m.
• E. 11:30 p.m.

Hint:

When setting expressions for distance equal, remember that distance is always positive. This means you should solve for both \(A=B\) and \(A=-B\). If multiple positive times result, the first one is usually the intended answer unless the question specifies otherwise.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem provides two linear functions representing the distance of two trains from a central point. The term "equidistant" means that their distances are equal. Since distance is a non-negative quantity, we must set the absolute values of the two expressions equal to each other to find the time \(t\).
Step 2: Key Formula or Approach:
To find when the trains are equidistant, we solve the equation: \[ |-10t + 115| = |-20t + 150| \] This absolute value equation leads to two separate linear equations: 1. \(-10t + 115 = -20t + 150\) 2. \(-10t + 115 = -(-20t + 150)\) We solve for \(t\) in each case and choose the first valid, positive time.
Step 3: Detailed Explanation:
Case 1: The expressions are equal. This corresponds to the trains being on the same side of Centerville. \[ -10t + 115 = -20t + 150 \] Add \(20t\) to both sides of the equation: \[ 10t + 115 = 150 \] Subtract 115 from both sides: \[ 10t = 35 \] \[ t = 3.5 \] A time of \(t = 3.5\) hours after 12:00 noon is 3:30 p.m. This is a valid solution. Case 2: The expressions are opposites. This corresponds to the trains being on opposite sides of Centerville. \[ -10t + 115 = -(-20t + 150) \] \[ -10t + 115 = 20t - 150 \] Add \(10t\) to both sides: \[ 115 = 30t - 150 \] Add 150 to both sides: \[ 265 = 30t \] \[ t = \frac{265}{30} = \frac{53}{6} \approx 8.833 \text{ hours} \] To convert this to minutes: \(0.833 \times 60 \approx 50\) minutes. This corresponds to 8:50 p.m. Since both 3:30 p.m. and 8:50 p.m. are options, and the question asks for "what time", the convention is to provide the first time the event occurs. Step 4: Final Answer:
The first time the trains are equidistant is at \(t=3.5\) hours, or 3:30 p.m.