A train covers a certain distance at a speed of 240 km/h in 5 hours. If a flight has to cover the same distance in 45 mins, it must travel at a speed of:
1600 km/h
This problem involves the concepts of speed, distance, and time. We are given the speed and time for a train journey and asked to find the speed a flight needs to travel at to cover the same distance in a different amount of time. The key is to first find the distance covered, which remains constant for both the train and the flight.
The train travels at a certain speed for a given time. The relationship between speed, distance, and time is:
\(\text{Distance} = \text{Speed} \times \text{Time}\)
Given:
Let's calculate the distance covered by the train:
\(\text{Distance} = 240 \text{ km/h} \times 5 \text{ hours}\)
\(\text{Distance} = 1200 \text{ km}\)
So, the total distance covered is 1200 km.
The flight needs to cover the same distance in 45 minutes. To use the speed formula with speed in km/h, we need the time in hours. There are 60 minutes in an hour.
\(\text{Flight Time} = 45 \text{ minutes}\)
Convert minutes to hours:
\(\text{Flight Time in hours} = \frac{45 \text{ minutes}}{60 \text{ minutes/hour}}\)
\(\text{Flight Time in hours} = \frac{45}{60} \text{ hours}\)
\(\text{Flight Time in hours} = \frac{3}{4} \text{ hours} = 0.75 \text{ hours}\)
The flight has to cover the 1200 km distance in 0.75 hours.
Now we need to find the speed the flight must travel at to cover the 1200 km distance in 0.75 hours. We use the same formula, rearranged to solve for speed:
\(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)
Given:
Calculate the required flight speed:
\(\text{Flight Speed} = \frac{1200 \text{ km}}{0.75 \text{ hours}}\)
\(\text{Flight Speed} = \frac{1200}{3/4} \text{ km/h}\)
\(\text{Flight Speed} = 1200 \times \frac{4}{3} \text{ km/h}\)
\(\text{Flight Speed} = \frac{1200 \times 4}{3} \text{ km/h}\)
\(\text{Flight Speed} = \frac{4800}{3} \text{ km/h}\)
\(\text{Flight Speed} = 1600 \text{ km/h}\)
Therefore, the flight must travel at a speed of 1600 km/h to cover the same distance in 45 minutes.
| Vehicle | Speed | Time | Distance |
|---|---|---|---|
| Train | 240 km/h | 5 hours | \(240 \times 5 = 1200\) km |
| Flight | ? | 45 mins (0.75 hours) | 1200 km |
Required Flight Speed \( = \frac{\text{Distance}}{\text{Time}} = \frac{1200 \text{ km}}{0.75 \text{ hours}} = 1600 \text{ km/h}\).
The calculated speed for the flight is 1600 km/h.
| Concept | Formula | Units (Example: km, hours) |
|---|---|---|
| Distance | \(\text{Distance} = \text{Speed} \times \text{Time}\) | km = (km/h) \(\times\) (h) |
| Speed | \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\) | km/h = km / h |
| Time | \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\) | h = km / (km/h) |
It is crucial to ensure that units are consistent when performing calculations involving speed, distance, and time. If speed is in km/h, time should be in hours, and distance in km. If units are mixed (e.g., speed in km/h and time in minutes), one of them must be converted to match the required unit system.
In this problem, converting the flight time from minutes to hours was an essential step to correctly apply the speed-distance-time formula with speed given in km/h.
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