Question

A train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

Short answer

Answer: The moving train will be approximately 33.3 meters away from the stationary train when it comes to a stop.

Step-by-step solution

Find time taken for the moving train to stop

To find out the time it takes for the moving train to come to a stop, we will use the following equation of motion: \(v = u + at\) Where: \(v\) - final velocity (0 m/s since the train comes to a stop) \(u\) - initial velocity (40.0 m/s, given) \(a\) - acceleration (-6.0 m/s², since it is decelerating) \(t\) - time (the unknown we are solving for) Now, let's plug in the given values into the equation and solve for the time `t`. \(0 = 40.0 + (-6.0) \cdot t\)

Solve for time 't'

Solving for t in the equation: \(-40.0 = -6.0t\) \(t = \frac{-40.0}{-6.0}\) \(t = 6.\bar{6} \ \text{seconds}\)

Calculate the distance traveled by the moving train

Now, we will use another equation of motion to find the distance covered by the moving train during the time it took to come to a stop. \(s = ut + \frac{1}{2}at^2\) Where: \(s\) - distance traveled (unknown we are solving for) \(u\) - initial velocity (40.0 m/s, given) \(a\) - acceleration (-6.0 m/s², since it is decelerating) \(t\) - time (6.\bar{6} seconds, found in step 2) Now let's plug in the values into the equation and solve for the distance `s`. \(s = 40.0 \cdot 6.\bar{6} + \frac{1}{2} \cdot (-6.0) \cdot (6.\bar{6})^2\)

Solve for distance 's'

Solving for s in the equation: \(s = 40.0 \cdot 6.\bar{6} - 3.0 \cdot (6.\bar{6})^2\) \(s \approx 133.3 \ \text{meters}\)

Calculate the distance between the stationary train and the moving train when it stops

We now know that the moving train traveled approximately 133.3 meters before coming to a stop. Since the initial distance between the trains was 100 meters, we need to calculate the difference to find out how far the moving train is from the stationary train when it stops. Distance between the trains when moving train stops = Distance traveled by moving train - Initial distance between trains \(\approx 133.3 - 100.0 = 33.3 \ \text{meters}\) So, the moving train will be approximately 33.3 meters away from the stationary train when it comes to a stop.

Key concepts

Equations of Motion

In physics, the equations of motion are critical tools for determining how objects move under the influence of various forces. They describe the relationship between displacement, velocity, acceleration, and time. There are three primary equations, but in problems like the one we examined, we often use two key formulas:

• First, the equation for final velocity: \(v = u + at\)
• Second, the equation for displacement: \(s = ut + \frac{1}{2}at^2\)

Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, \(t\) is the time covered, and \(s\) is the displacement. Applying these formulas helps us solve real-world problems like stopping distances or collision avoidance. Understanding how to substitute the known values into these equations can unravel much about motion's characteristics.

Deceleration

Deceleration refers to the reduction of speed or slowing down process, essentially a negative acceleration. In our equation of motion, acceleration \(a\) is replaced by negative values when describing deceleration. Applying this to our train scenario, since it decelerates at \(-6.0 \text{ m/s}^2\), a constant braking force acts against its motion, causing it to stop.

Deceleration is particularly important in safety analyses, such as determining how fast a vehicle can stop to avoid a crash. By understanding deceleration, engineers can design more effective braking systems. In calculations, always keep in mind that deceleration is a vector quantity, meaning it has direction and magnitude.

Relative Motion

Relative motion is the concept of viewing the motion of one object with reference to another. In this problem, we consider the motion of a moving train concerning a stationary train. Initially, the stationary train is 100 meters away from the moving train. Relative motion enables us to analyze their positions concerning each other as time passes.

Since one train is at rest, it simplifies our calculations of relative position by directly using the moving train's displacement. Understanding relative motion is essential in aligning and predicting paths in traffic systems, collision avoidance, and navigational systems.

Initial Velocity

The initial velocity \(u\) is the speed of an object when observation starts. In our problem, it is the velocity at which the moving train begins its journey towards the other stationary train and is given as \(40.0 \text{ m/s}\).

Initial velocity is critical when predicting an object's subsequent motion and understanding how external forces like deceleration affect its journey. Knowing \(u\) gives us a starting point to calculate how long it would take to stop the object or reach a particular destination, just like finding out the stopping distance in our problem.

Final Velocity

Final velocity \(v\) refers to the speed of an object at the end of the observation period. For the moving train, the final velocity is \(0 \text{ m/s}\) because it comes to a stop. This zero value is crucial for problem-solving as it sets the condition for calculating how long it has taken for the train to reach a halt or how far it traveled during deceleration.

Knowing the final velocity helps verify outcomes, ensuring that the solution aligns with the known facts of the scenario. It also provides insight into the efficacy of any deceleration factors present in the system.