Calculating Acceleration of a Train
Understanding the principles of motion is crucial for both aspiring engineers and railway enthusiasts. One fundamental aspect of motion is acceleration, which is pivotal in railway operations. In this post, we will solve a classic problem: finding the acceleration of a train starting from rest and reaching a speed of 40 km/h in 10 minutes. This example will not only clarify the process but also enhance your problem-solving skills. Additionally, we will explore similar problems to give you a broader perspective. For more insights into railway physics and mathematics, visit Railway Study.
Problem Statement
A train starts from a railway station with uniform acceleration and reaches a speed of 40 km/h in 10 minutes. We need to find the acceleration of the train.
Understanding the Problem
- Initial Velocity (u): The train starts from rest, so the initial velocity u=0u = 0u=0.
- Final Velocity (v): The train attains a speed of 40 km/h. To use standard units in calculations, convert this speed into meters per second (m/s).40 km/h=40×10003600 m/s=400003600 m/s≈11.11 m/s40 text{ km/h} = frac{40 times 1000}{3600} text{ m/s} = frac{40000}{3600} text{ m/s} approx 11.11 text{ m/s}40 km/h=360040×1000 m/s=360040000 m/s≈11.11 m/s
- Time (t): The time taken to reach this speed is 10 minutes, which needs to be converted into seconds.10 minutes=10×60 seconds=600 seconds10 text{ minutes} = 10 times 60 text{ seconds} = 600 text{ seconds}10 minutes=10×60 seconds=600 seconds
Finding the Acceleration
To find the acceleration aaa, use the formula for uniform acceleration:v=u+atv = u + atv=u+at
Rearrange to solve for aaa:a=v−uta = frac{v – u}{t}a=tv−u
Substitute the known values:a=11.11−0600≈11.11600≈0.0185 m/s2a = frac{11.11 – 0}{600} approx frac{11.11}{600} approx 0.0185 text{ m/s}^2a=60011.11−0≈60011.11≈0.0185 m/s2
Conclusion
The acceleration of the train is approximately 0.0185 m/s20.0185 text{ m/s}^20.0185 m/s2. This calculation demonstrates the basic principle of motion under uniform acceleration. For more detailed explanations and practice problems, visit Railway Study.
Similar Problems and Solutions
1. Problem: Acceleration of a Car in Traffic
A car starts from rest and accelerates uniformly to a speed of 60 km/h in 5 minutes. Find the acceleration.
Solution:
- Convert Speed:60 km/h=60×10003600 m/s=600003600 m/s≈16.67 m/s60 text{ km/h} = frac{60 times 1000}{3600} text{ m/s} = frac{60000}{3600} text{ m/s} approx 16.67 text{ m/s}60 km/h=360060×1000 m/s=360060000 m/s≈16.67 m/s
- Convert Time:5 minutes=5×60 seconds=300 seconds5 text{ minutes} = 5 times 60 text{ seconds} = 300 text{ seconds}5 minutes=5×60 seconds=300 seconds
- Calculate Acceleration:a=16.67−0300≈16.67300≈0.0556 m/s2a = frac{16.67 – 0}{300} approx frac{16.67}{300} approx 0.0556 text{ m/s}^2a=30016.67−0≈30016.67≈0.0556 m/s2
2. Problem: Acceleration of an Elevator
An elevator starts from rest and reaches a speed of 2 m/s in 8 seconds. Determine its acceleration.
Solution:
- Initial Velocity (u): 0 m/s
- Final Velocity (v): 2 m/s
- Time (t): 8 seconds
- Calculate Acceleration:a=2−08=28=0.25 m/s2a = frac{2 – 0}{8} = frac{2}{8} = 0.25 text{ m/s}^2a=82−0=82=0.25 m/s2
3. Problem: Acceleration of a Bicycle
A bicycle accelerates from 5 m/s to 15 m/s in 10 seconds. Find the acceleration.
Solution:
- Initial Velocity (u): 5 m/s
- Final Velocity (v): 15 m/s
- Time (t): 10 seconds
- Calculate Acceleration:a=15−510=1010=1 m/s2a = frac{15 – 5}{10} = frac{10}{10} = 1 text{ m/s}^2a=1015−5=1010=1 m/s2
Key Takeaways
- Unit Conversion: Always convert speeds to meters per second and time to seconds for uniform calculations.
- Formula Application: Use the formula a=v−uta = frac{v – u}{t}a=tv−u to find acceleration under uniform conditions.
- Practice: Solve various problems to become proficient in applying these concepts.
For more detailed explanations and a wide range of problems related to railway physics, motion, and acceleration, check out Railway Study. Our platform offers a wealth of resources to help you master these essential concepts.
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