Problem 92 - Fundamentals of Physics

Question

Two subway stops are separated by \(1100 \mathrm{~m}\). If a subway train accelerates at \(+1.2 \mathrm{~m} / \mathrm{s}^{2}\) from rest through the first hall of the distance and decelerates at \(-1.2 \mathrm{~m} / \mathrm{s}^{2}\) through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph \(x, v\) and \(a\) versus \(t\) for the trip.

Step-by-Step Solution

Answer

Travel time is 60.56 s and maximum speed is 36.33 m/s.

Step 1: Understand the problem

The subway train accelerates from rest at a constant acceleration for half of the total distance and decelerates at the same constant rate for the other half. We need to find the total travel time and maximum speed, and graph the position, velocity, and acceleration over time.

Step 2: Determine Distance for Each Phase

The total distance between the two stops is 1100 m. Since the train accelerates and decelerates equally over the trip, each phase covers half the distance: \[ d_1 = \frac{1100}{2} \text{ m} = 550 \text{ m} \] where \(d_1\) is the distance for each phase (acceleration and deceleration).

Step 3: Use Kinematic Equation for Acceleration Phase

We use the kinematic equation for the phase of acceleration: \( v^2 = u^2 + 2as_1 \). Here, initial velocity \( u = 0 \), \( a = 1.2 \text{ m/s}^2 \), and \( s_1 = 550 \text{ m} \). Solving for \(v\) (maximum speed): \[ v^2 = 0 + 2 \times 1.2 \times 550 \] \[ v = \sqrt{1320} \] \[ v \approx 36.33 \text{ m/s} \]

Step 4: Calculate Time for Each Phase

From the acceleration phase, use \( v = u + at \) to find time \( t_1 \) for the acceleration phase: \[ t_1 = \frac{v}{a} = \frac{36.33}{1.2} \approx 30.28 \text{ s} \]. The deceleration phase is symmetrical, so it takes the same amount of time: \( t_2 = 30.28 \text{ s} \).

Step 5: Calculate Total Travel Time

The total time is the sum of the time for both phases, since the motion is symmetrical: \[ t = t_1 + t_2 = 30.28 + 30.28 \approx 60.56 \text{ s} \].

Step 6: Graph Position, Velocity, and Acceleration versus Time

- **Position (x vs. t)**: Starts at 0, increases parabolically to 550 m at 30.28 s, then symmetrically to 1100 m at 60.56 s. - **Velocity (v vs. t)**: Starts at 0, linearly increases to maximum 36.33 m/s at 30.28 s, then linearly decreases back to 0 at 60.56 s. - **Acceleration (a vs. t)**: Constant at +1.2 m/s² for 30.28 s, then constant at -1.2 m/s² for next 30.28 s.

Key Concepts

Acceleration

Acceleration is a crucial concept in kinematics, especially when analyzing the motion of a body under increasing speed. It is defined as the rate of change of velocity over time. In this scenario, the subway train starts its journey from rest, which means it has an initial velocity of zero. However, as it accelerates at a constant rate of

• 1.2 m/s²

through the first half of its trip, it picks up speed progressively. The rate of acceleration determines how quickly the train reaches its maximum speed during the first phase of its journey. Understanding acceleration helps predict the motion behavior of moving objects conveniently.

Kinematic Equations

Kinematic equations are powerful tools in physics that describe the motion of objects when their acceleration is constant. These equations allow us to connect different motion parameters such as displacement, initial velocity, final velocity, acceleration, and time. In this exercise:

• The equation: \( v^2 = u^2 + 2as \) helps find the maximum speed \( v \) reached by the train when accelerating.
• Another equation: \( v = u + at \) helps calculate the time \( t \) taken to reach that speed.

By strategically using these equations, you can break down the motion into comprehensible parts, calculate each phase, and find the desired results precisely.

Maximum Speed

The maximum speed of the subway train is attained at the end of the acceleration phase. Since both phases (acceleration and deceleration) cover half of the total distance (550 meters each), the maximum speed is reached at the midpoint of the journey before starting the deceleration. Using the kinematic equation \( v^2 = 2as \) with

• \( a = 1.2 \text{ m/s}^2 \)
• \( s = 550 \text{ m} \)

We deduce that the maximum speed \( v \) is approximately 36.33 m/s. This maximum speed indicates the peak of the velocity graph and is achieved exactly at the halfway mark, where the train begins to decelerate.

Deceleration

Deceleration, or negative acceleration, is equally significant as acceleration when analyzing motion. It occurs when a body slows down, reducing its speed. In this scenario, the train starts decelerating at the midpoint of its journey at a constant rate of

• -1.2 m/s²

The deceleration phase is mirrored after the acceleration phase. As a result, the train's velocity declines linearly over the same distance until it comes to a stop just before reaching the next station. The amount of time taken for the train to completely decelerate is the same as the acceleration time, due to the identical magnitude of acceleration and deceleration.