Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

1 answer

Let the distance from Mr. Earl E. Bird's house to his workplace be denoted by \( d \) miles.

We start by analyzing the two scenarios provided:

  1. When traveling at 40 mph:

    • The time taken to travel the distance \( d \) is given by: \[ \text{Time} = \frac{d}{40} \text{ hours} \]
    • Since Mr. Bird arrives 3 minutes late, he should have arrived at 8:00 A.M. + \( t \) to be on time, where \( t \) is the time he would take to arrive on time in hours.
    • Converting 3 minutes to hours gives \( \frac{3}{60} = 0.05 \) hours.
    • Thus, we have: \[ \frac{d}{40} = t + 0.05 \tag{1} \]
  2. When traveling at 60 mph:

    • The time taken at this speed is: \[ \text{Time} = \frac{d}{60} \text{ hours} \]
    • In this scenario, he arrives 3 minutes early, meaning: \[ \frac{d}{60} = t - 0.05 \tag{2} \]

Now we have two equations (1) and (2): \[ \frac{d}{40} = t + 0.05 \] and \[ \frac{d}{60} = t - 0.05 \]

Next, we solve for \( t \) in both equations: From equation (1): \[ t = \frac{d}{40} - 0.05 \] From equation (2): \[ t = \frac{d}{60} + 0.05 \]

Setting the expressions for \( t \) equal to each other: \[ \frac{d}{40} - 0.05 = \frac{d}{60} + 0.05 \]

To eliminate the fractions, we can multiply the entire equation by 120 (the least common multiple of 40 and 60): \[ 120 \left( \frac{d}{40} \right) - 120 \cdot 0.05 = 120 \left( \frac{d}{60} \right) + 120 \cdot 0.05 \]

This simplifies to: \[ 3d - 6 = 2d + 6 \]

Now, let's isolate \( d \): \[ 3d - 2d = 6 + 6 \] \[ d = 12 \]

Now that we have the distance \( d \) (12 miles), we can calculate the time \( t \) it should take to reach on time: Using equation for \( t \): \[ t = \frac{d}{40} - 0.05 = \frac{12}{40} - 0.05 = 0.3 - 0.05 = 0.25 \text{ hours} \] This means Mr. Bird needs to arrive at his workplace in 0.25 hours.

To find the average speed \( s \) at which he should drive to arrive on time, we use the formula: \[ s = \frac{d}{t} = \frac{12 \text{ miles}}{0.25 \text{ hours}} = 48 \text{ miles per hour} \]

Thus, the average speed at which Mr. Bird should drive to arrive at his workplace precisely on time is: \[ \boxed{48} \text{ miles per hour} \]