A train runs at a speed of 90 km/h in the first 10 minutes and 60 km/h in next 25 minutes and 15 km/h in last 4 minutes. What is the average speed of the train?
The question asks for the average speed of a train that travels at different speeds over different time intervals. To find the average speed, we need to calculate the total distance covered by the train and divide it by the total time taken for the journey.
The formula for average speed is:
$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$
The journey is divided into three segments with different speeds and durations.
We know that Distance = Speed $\times$ Time. We must ensure that units are consistent. The speeds are given in km/h, and the times are given in minutes. Let's convert the time for each segment into hours.
The total distance is the sum of the distances covered in each segment.
$$ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3 $$
$$ \text{Total Distance} = 15 \text{ km} + 25 \text{ km} + 1 \text{ km} = 41 \text{ km} $$
The total time is the sum of the time taken for each segment. It's easier to sum in minutes first and then convert to hours for the final calculation.
Total time in minutes = 10 minutes + 25 minutes + 4 minutes = 39 minutes.
Converting total time to hours:
$$ \text{Total Time} = 39 \text{ minutes} = 39 \times \frac{1}{60} \text{ hours} = \frac{39}{60} \text{ hours} = \frac{13}{20} \text{ hours} $$
Now, use the formula for average speed with the total distance and total time.
$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{41 \text{ km}}{\frac{13}{20} \text{ h}} $$
$$ \text{Average Speed} = 41 \times \frac{20}{13} \text{ km/h} = \frac{820}{13} \text{ km/h} $$
To convert $\frac{820}{13}$ to a mixed fraction, divide 820 by 13.
$$ 820 \div 13 $$
$$ 820 = 13 \times 63 + 1 $$
The quotient is 63 and the remainder is 1. So, the mixed fraction is $63 \frac{1}{13}$.
Therefore, the average speed of the train is $63 \frac{1}{13}$ km/h.
| Segment | Speed (km/h) | Time (minutes) | Time (hours) | Distance (km) |
|---|---|---|---|---|
| 1 | 90 | 10 | $10/60 = 1/6$ | $90 \times 1/6 = 15$ |
| 2 | 60 | 25 | $25/60 = 5/12$ | $60 \times 5/12 = 25$ |
| 3 | 15 | 4 | $4/60 = 1/15$ | $15 \times 1/15 = 1$ |
| Total | - | 39 | $39/60 = 13/20$ | $15 + 25 + 1 = 41$ |
Average Speed = Total Distance / Total Time = $41 \text{ km} / (13/20) \text{ h} = 41 \times 20/13 \text{ km/h} = 820/13 \text{ km/h}$
$$ \frac{820}{13} = 63 \frac{1}{13} $$
The average speed of the train is $63 \frac{1}{13}$ km/h.
| Concept | Formula/Definition | Application in this problem |
|---|---|---|
| Average Speed | Total Distance / Total Time | Used to find the overall speed considering varying speeds. |
| Distance | Speed $\times$ Time | Calculated for each segment of the journey. |
| Unit Conversion | Minutes to Hours (divide by 60) | Essential for consistency when speed is in km/h. |
| Total Distance | Sum of distances of all segments | Added $15 \text{ km} + 25 \text{ km} + 1 \text{ km} = 41 \text{ km}$. |
| Total Time | Sum of time for all segments | Added $10 + 25 + 4 = 39$ minutes, converted to $13/20$ hours. |
Speed, distance, and time are fundamental concepts in physics and mathematics, often related by the formula: $$ \text{Speed} = \frac{\text{Distance}}{\text{Time}} $$ This can be rearranged to find distance ($\text{Distance} = \text{Speed} \times \text{Time}$) or time ($\text{Time} = \frac{\text{Distance}}{\text{Speed}}$).
When an object travels at different speeds over different parts of its journey, the average speed is not simply the average of the speeds. It must be calculated using the total distance and the total time.
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