Answer:
Option 2nd is correct
[tex]50(x-0.5)[/tex] miles
Step-by-step explanation:
Using distance formula:
[tex]\text{Distance} = \text{Speed} \times \text{time}[/tex]
As per the statement:
A train traveling 50 mph left a station 30 minutes before a second train running at 55 mph. If 55x represents the distance the faster train travels.
Second train data:
Let time taken by second train be x hrs
Speed = 55 mph
then;
Distance = 55x miles
First train data:
Speed = 50 mph
time = [tex]x - \frac{30}{60} = x - \frac{1}{2}[/tex]
then;
[tex]\text{Distance} = 50(x-\frac{1}{2})[/tex] miles
Since, the second train travels faster than first train
We have to find the distance of the slower train.
Distance of the slower train = [tex]50(x-0.5)[/tex] miles
Therefore, the following algebraic expressions represents the distance of the slower train is, [tex]50(x-0.5)[/tex]
