Calculating Distance Traveled by a Train under Constant Acceleration

Consider a train that starts from rest and accelerates at a rate of 1 m/s2 for 10 seconds. How can we determine the distance it travels during this acceleration period?

Using Kinematic Equations to Solve the Problem

To find the distance traveled by the train, we can utilize the following kinematic equation:

d v_i t frac{1}{2} a t^2

Here:

d is the distance traveled, v_i is the initial velocity, which is 0 m/s as the train starts from rest, a is the acceleration, which is given as 1 m/s2, t is the time, which is 10 seconds.

Substituting these values into the equation, we have:

d 0 cdot 10 frac{1}{2} cdot 1 cdot 10^2

Thus:

d 0 frac{1}{2} cdot 100

d frac{100}{2} 50 text{ meters}

Understanding the Intuition

The above mathematical approach is effective, but it might not provide a clear understanding of why the train travels 50 meters. Let's explore the underlying intuition:

For every second of motion, the speed of the train increases by 1 m/s. Therefore, the speed can be represented as follows:

At t 1, v 1 m/s,

At t 2, v 2 m/s,

…

At t 10, v 10 m/s.

If we consider the average speed over the 10 seconds, it would be the average of the initial speed (0 m/s) and the final speed (10 m/s), which is:

frac{0 10}{2} 5 text{ m/s}

Given that the train averages 5 meters per second for 10 seconds, the total distance traveled is:

5 text{ m/s} times 10 text{ s} 50 text{ meters}

Visualizing with a Graph

To further understand the relationship between time and the distance traveled, you can draw a graph with time along the x-axis and speed on the y-axis. The distance traveled will be represented by the area under the graph. For example:

2 seconds at 3 m/s would be represented as 3 x 2 6 meters.

For this problem, the area under the line (a triangle) would be:

0.5 times 10 text{s} times 10 text{m/s} 50 text{ meters}

Final Thoughts

Therefore, the train travels a distance of 50 meters in 10 seconds of uniform acceleration.

Key Concepts:

The equation S ut frac{1}{2}at^2 is useful for calculating displacement (S) given initial velocity (u), acceleration (a), and time (t). The method of calculating distance by finding the area under the velocity-time graph is particularly intuitive.

Feel free to ask if you have any other questions or need further clarification!