Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Starting at station A, a commuter train accelerates at 3 meters per second per second for 8 seconds, then travels at constant speed \(v_{m}\) for 100 seconds, and finally brakes (decelerates) to a stop at station \(B\) at 4 meters per second per second. Find (a) \(v_{m}\) and (b) the distance between \(A\) and \(B\).

Step-by-Step Solution

Answer

(a) \( v_m = 24 \text{ m/s} \), (b) Distance = 2568 m.

Step 1: Understand the Problem

To solve the problem, we need to determine two quantities: the maximum speed \( v_m \) reached, and the total distance traveled from station \( A \) to station \( B \). The journey consists of three phases: acceleration, constant speed travel, and deceleration.

Step 2: Calculate Maximum Speed v_m

During the acceleration phase, the train accelerates at \( 3 \text{ m/s}^2 \) for \( 8 \text{ s} \). Using the formula for final velocity in uniform acceleration, \( v = u + at \), where \( u = 0 \), \( a = 3 \text{ m/s}^2 \), and \( t = 8 \text{ s} \), we find:\[ v_m = 0 + 3 \times 8 = 24 \text{ m/s} \].

Step 3: Travel at Constant Speed

During this phase, the train travels at the constant speed \( v_m = 24 \text{ m/s} \) for \( 100 \text{ s} \). The distance covered during this time is given by \( s = vt \), which calculates to:\[ s_2 = 24 \times 100 = 2400 \text{ m} \].

Step 4: Calculate Stopping Distance

The train decelerates at \( 4 \text{ m/s}^2 \) until it stops. Using the formula \( v^2 = u^2 - 2as \), where \( v = 0 \), \( u = 24 \text{ m/s} \), and \( a = 4 \text{ m/s}^2 \), we solve for \( s \):\[ 0 = 24^2 - 2 \times 4 \times s_3 \] resulting in \[ s_3 = \frac{24^2}{2 \times 4} = 72 \text{ m} \].

Step 5: Calculate Total Distance

Add the distances from each phase: acceleration, constant-speed travel, and deceleration. From Step 2, during acceleration \( 24 \text{ m/s}^2 * 8 \text{ s} = 96 \text{ m} \); from Step 3, \( 2400 \text{ m} \); and from Step 4, \( 72 \text{ m} \). The total distance \( AB \) is the sum:\[ 96 + 2400 + 72 = 2568 \text{ m} \].

Key Concepts

Understanding Kinematics

Kinematics is a branch of classical mechanics that deals with the motion of objects without considering the causes of motion. It focuses on parameters like displacement, velocity, and acceleration. When studying the motion of a train between two stations, kinematics allows us to understand how the train's speed and position change over time. This involves analyzing its acceleration, constant velocity, and how it slows down.

Key elements in kinematics involve:

• Position: Where the object is located at a certain time.
• Velocity: How fast the object is moving and in what direction.
• Acceleration: The rate at which the object's velocity changes.

Uniform Acceleration

Uniform acceleration is when an object's velocity changes at a constant rate. For the train leaving station A, it begins with a uniform acceleration of 3 meters per second squared for 8 seconds.

This means every second the train's speed increases by 3 meters per second. Starting from rest, the train reaches a speed of 24 meters per second by the end of the acceleration period.
The formula for this is
\[ v = u + at \]
where:

• \( u \) is the initial velocity (0 m/s in this case).
• \( a \) is the acceleration (3 m/s²).
• \( t \) is the time (8 seconds).

By substituting these values in, we confirm that the velocity at the end of this phase is indeed 24 m/s.

Constant Velocity

After the train reaches its maximum speed of 24 meters per second, it travels at this constant velocity for 100 seconds. Constant velocity means that the train's speed and direction do not change during this time.

This phase of the journey is straightforward, as there is no acceleration affecting the train's speed. The formula used to calculate the distance traveled while moving at a constant speed is:
\[ s = vt \]
where:

• \( s \) is the distance traveled.
• \( v \) is the velocity (24 m/s).
• \( t \) is the time (100 seconds).

Thus, the distance covered at constant speed is 2400 meters.

Deceleration

Deceleration occurs when an object slows down. In this problem, the train decelerates at 4 meters per second squared as it approaches station B.

Deceleration is essentially negative acceleration, meaning the speed of the train decreases over time. During this phase, the train needs to come to a complete stop from 24 meters per second.
To find the stopping distance, we use:
\[ v^2 = u^2 - 2as \]
where:

• \( v \) is the final velocity (0 m/s, because the train stops).
• \( u \) is the initial velocity (24 m/s).
• \( a \) is the deceleration (4 m/s²).
• \( s \) is the stopping distance.

Using these values, we find that the stopping distance is 72 meters.