To find the speed of the train, we need to know two things: the distance it travels and the time it takes to cover that distance. In this problem, we are given the lengths of two stations and the time it takes for the train to pass each station.
We'll start with the first station. Let's denote the length of the train as \( L_t \) and its speed as \( v \). As the train passes the station, it covers the length of the station plus its own length, so the total distance it covers while passing the first station is \( 450 + L_t \) meters. It takes 27 seconds to do this, so we can express its speed as:
[tex]\[v = \frac{{450 + L_t}}{27}\][/tex]
Next, we consider the second station. The total distance the train covers while passing this station is [tex]\( 330 + L_t \)[/tex], and it takes 21 seconds to pass it. Similarly, the speed of the train is:
[tex]\[v = \frac{{330 + L_t}}{21}\][/tex]
Because the train is running at a uniform speed, the speed \( v \) is the same in both cases. Therefore, we can set the two expressions for \( v \) equal to each other:
[tex]\[\frac{{450 + L_t}}{27} = \frac{{330 + L_t}}{21}\][/tex]
To find \( L_t \), we solve this equation for \( L_t \):
[tex]\[21 \cdot (450 + L_t) = 27 \cdot (330 + L_t)\][/tex]
Multiplying out the terms, we get:
[tex]\[9450 + 21L_t = 8910 + 27L_t\][/tex]
Subtract[tex]\( 21L_t \)[/tex]from both sides to get the terms involving \( L_t \) on one side:
[tex]\[9450 = 8910 + 6L_t\][/tex]
Subtract \( 8910 \) from both sides:
[tex]\[540 = 6L_t\][/tex]
Divide by 6 to solve for \( L_t \):
[tex]\[L_t = \frac{540}{6} = 90\][/tex]
The length of the train is 90 meters.
Now that we know \( L_t \), we can find the speed \( v \) using either equation for the speed. We'll use the first one (it doesn't matter which one you choose):
[tex]\[v = \frac{{450 + L_t}}{27} = \frac{{450 + 90}}{27} = \frac{540}{27} = 20 \text{ meters/second}\][/tex]
The speed of the train is 20 meters per second. If you want to convert this to kilometers per hour (km/h), recall that [tex]\( 1 \text{ m/s} = 3.6 \text{ km/h} \)[/tex], so:
[tex]\[v = 20 \text{ m/s} \times 3.6 \frac{\text{km}}{\text{h per m/s}} = 72 \text{ km/h}\][/tex]
Therefore, the speed of the train is 20 meters per second or 72 kilometers per hour.
