Question

Starting from rest, a bus increases speed at constant acceleration \(a_{1}\), then travels at constant speed \(v_{m}\), and finally brakes to a stop at constant acceleration \(a_{2}\left(a_{2}<0\right) .\) It took 4 minutes to travel the 2 miles between stop \(\mathrm{C}\) and stop \(\mathrm{D}\) and then 3 minutes to go the \(1.4\) miles between stop \(\mathrm{D}\) and stop \(\mathrm{E}\). (a) Sketch the graph of the velocity \(v\) as a function of time \(t\), \(0 \leq t \leq 7\) (b) Find the maximum speed \(v_{m-}\) (c) If \(a_{1}=-a_{2}=a\), evaluate \(a\).

Short answer

Maximum speed is determined using uniform condition blending acceleration equations, and acceleration is evaluated as equal magnitude parts in different intervals.

Step-by-step solution

Understand Initial Conditions and Units

Before solving the problem, we note that the bus travels the distance between stops C and D in 4 minutes and between stops D and E in 3 minutes. The total distance traveled from C to E is 2 miles + 1.4 miles = 3.4 miles. Also, we note that time is given in minutes and distance in miles, so we will need to convert them later if necessary.

Analyzing the Graph Sketch Requirement

The velocity-time graph involves three segments: acceleration from rest, constant speed, and deceleration to rest. Since the problem states that the bus first accelerates, reaches a constant speed, and then brakes to zero velocity, the graph must first rise linearly, then be a horizontal line, and finally fall linearly back to zero. Label each segment: - From t=0 to t=t1, velocity increases linearly. - From t=t1 to t=t2, velocity is constant at maximum speed v_m. - From t=t2 to t=7 minutes, velocity decreases linearly.

Determine Maximum Velocity for Part (b)

To find the maximum speed, set up and solve the equations for the distances traveled. Given total travel times and distances, use:- Equation 1: The distance covered at maximum speed: \( v_{m} \times (t_2 - t_1) \)- Equation 2: Distance covered during acceleration is same during deceleration: \( \frac{1}{2} a_1 t_1^2 \), substitute for \( a_1 \). Since the bus accelerates to the same speed it decelerates from, and given the unification of travel in both segments, solve for \( v_{m} \).

Calculate Acceleration for Part (c)

Given \( a_1 = -a_2 = a \), use the equations from step 3. Plug-in the required constant acceleration and solve the distance traversed during each interval equation. Ensure to convert miles to equivalent velocity or time units if needed for accuracy, like converting 4 min to hours or velocity into miles/min.

Key concepts

Velocity-Time Graph

A velocity-time (v-t) graph is one of the most important tools in motion analysis, especially when dealing with problems involving acceleration and velocity changes over time. To create a v-t graph, you need to understand how an object's speed evolves over time.
In our particular problem, the v-t graph encompasses three distinct segments:
1. **Acceleration Segment:** This is where the bus starts from rest and increases its speed. On the graph, this appears as a straight line moving upward, indicating a positive change in velocity over time.

• The slope of this line corresponds to the constant acceleration \(a_1\) of the bus.
• This phase continues until a time \(t_1\) when the bus reaches its maximum velocity \(v_m\).

2. **Constant Speed Segment:** When the bus travels at a constant speed, the graph is a horizontal line, reflecting no change in velocity. This happens between \(t_1\) and \(t_2\).
3. **Deceleration Segment:** Finally, the bus decelerates to a stop. This appears as a straight line moving downward on the graph, showing a negative change in velocity. This continues until the graph intercepts the time axis, indicating the bus has stopped. The slope here is equal to the negative acceleration \(a_2\).
Understanding each of these steps helps us draw an accurate picture of the motion described in the problem, displaying how the bus's velocity changes from start to stop.

Constant Acceleration

Constant acceleration occurs when an object's velocity changes at a steady rate over time. In this problem, constant acceleration plays a crucial role in determining how the bus moves between stops.
When the bus accelerates from a stop, its velocity increases linearly, meaning that every second, it gains the same amount of speed. This is represented in the equation:
\[v = u + at\]
Where:

• \(v\) is the final velocity,
• \(u\) is the initial velocity (which is zero since the bus starts from rest),
• \(a\) is the constant acceleration,
• \(t\) is time taken for the acceleration.

For this bus problem, constant acceleration occurs twice: once as it gains speed \(a_1\) and once as it slows down with \(a_2\) (where \(a_2 < 0\)). Both these acceleration phases play into finding out how fast the bus can go and how quickly it can stop.

Distance-Time Relationship

Understanding the relationship between distance and time is pivotal when analyzing motion in calculus. Calculus allows us to understand the nuances of this relationship.
For this problem, we calculate how far the bus travels during each phase by integrating the velocity function over time. The distance \(d\) traveled during a time interval can be calculated by:
\[d = \, v \, \times \left( t_{end} - t_{start} \right)\]
This equation is for the constant speed phase, represented as the area under the horizontal line on a v-t graph. For acceleration and deceleration phases, the area under the curve represents the distance, which is a triangle's area formula:
\[d = \, \frac{1}{2} \, a \, t^2\]
In essence, during acceleration, the bus covers increasing amounts of distance in equal intervals of time; when moving at a constant speed, distance grows linearly over time; and during deceleration, the distance it covers decreases until it stops, explaining the deceleration's symmetry in the problem.

Kinematics in Calculus

Kinematics in calculus offers a mathematical framework to analyze motion, allowing us to understand and predict the behavior of moving objects. We often deal with these primary kinematic equations:

• Converting velocity expressions between time and distance traveled.
• Evaluating the areas under the v-t graph, which equate to distances.

With the bus example, calculus is used to:
- Derive the velocity-time expressions for each travel phase.
- Integrate or evaluate the areas under these line segments to find cumulative distances.
Derivatives help us find velocity from position, while integrals help us find displacement from velocity. Given that \[s = \, \int v \, dt \]
and \[v = \, rac{ds}{dt},\]
calculus becomes essential for solving kinematics problems. It empowers us to not only sketch meaningful motion graphs but also quantify the motion's extent and limits. This harmonizes with our need to calculate precise accelerations \(a_1\) and \(a_2\) in the bus scenario, cementing calculus as an indispensable tool in analyzing kinematic systems.