Let's determine the freight train's average speed on the outbound trip. We can start by setting up some key details:
1. The outbound trip took 20 hours.
2. The return trip took 6 hours.
3. The speed on the return trip was 28 mph faster than the speed on the outbound trip.
Let's denote the speed of the train on the outbound trip as [tex]\( x \)[/tex] mph. Therefore, the speed on the return trip would be [tex]\( x + 28 \)[/tex] mph.
Since the distance between the two points is the same for both the outbound and return trips, we can establish the following relationships based on the formula for distance ([tex]\( \text{Distance} = \text{Speed} \times \text{Time} \)[/tex]):
For the outbound trip:
[tex]\[ \text{Distance} = x \times 20 \][/tex]
For the return trip:
[tex]\[ \text{Distance} = (x + 28) \times 6 \][/tex]
Since the distances are equal, we can set up the equation:
[tex]\[ x \times 20 = (x + 28) \times 6 \][/tex]
Now, solve for [tex]\( x \)[/tex]:
1. Expand the equation:
[tex]\[ 20x = 6x + 168 \][/tex]
2. Subtract [tex]\( 6x \)[/tex] from both sides to consolidate [tex]\( x \)[/tex] terms on one side:
[tex]\[ 20x - 6x = 168 \][/tex]
[tex]\[ 14x = 168 \][/tex]
3. Divide both sides by 14 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{168}{14} \][/tex]
[tex]\[ x = 12 \][/tex]
Therefore, the freight train's average speed on the outbound trip was 12 mph.
