Question

The distance from Centerville to a freight train is given by the expression \(-10t + 115\), and the distance from Centerville to a passenger train is given by the expression \(-20t + 150\). The expressions above give the distance from Centerville to each of two trains \( t \) hours after 12:00 noon. At what time after 12:00 noon will the trains be equidistant from Centerville?

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

Hint:

When two moving objects’ distances are given by linear expressions, equating those expressions gives the time when they are at the same distance.

Answer 1

The Correct Option is B

Step 1: Equating distances.
The freight train’s distance from Centerville after \( t \) hours: \[ d_f = -10t + 115 \] The passenger train’s distance from Centerville after \( t \) hours: \[ d_p = -20t + 150 \] For the trains to be equidistant from Centerville, we must have: \[ -10t + 115 = -20t + 150 \] Step 2: Solve for \( t \).
\[ -10t + 115 = -20t + 150 \] \[ 10t + 115 = 150 \] \[ 10t = 35 \quad \Rightarrow \quad t = 3.5 \] Step 3: Convert into time.
Since \( t = 3.5 \) hours = 3 hours and 30 minutes after noon, the time is 3:30.
Step 4: Conclusion.
The two trains will be equidistant from Centerville at: \[ \boxed{\text{(B) 3:30}} \]