91Ó°ÊÓ

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x\) -t graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

Step by step solution

01

Identify the Equation for Distance

The bus is accelerating and the student is running at a constant speed. The distance covered by the student is given by:\[d_s = v_s \cdot t\]where \(d_s\) is the distance covered by the student, \(v_s = 5.0\, \text{m/s}\) is the student's speed, and \(t\) is the time.The distance covered by the bus from the initial point of the student is:\[d_b = 40 + \frac{1}{2} a t^2\]where \(d_b\) is the distance covered by the bus, and \(a = 0.170\, \text{m/s}^2\) is the acceleration of the bus.

02

Set Up the Equation for the Intersection Point

To find when the student overtakes the bus, the distances both travel must be equal:\[v_s \cdot t = 40.0 + \frac{1}{2} a t^2\]Substitute the given values:\[5.0t = 40.0 + 0.085t^2\]

03

Solve the Quadratic Equation

Rearrange the equation to solve for \(t\):\[0.085t^2 - 5.0t + 40.0 = 0\]Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 0.085\), \(b = -5.0\), \(c = 40.0\), solve for \(t\):\[t = \frac{5.0 \pm \sqrt{(-5.0)^2 - 4 \times 0.085 \times 40.0}}{2 \times 0.085}\]Calculate the discriminant:\[t = \frac{5.0 \pm \sqrt{25 - 13.6}}{0.17}\]Solving further:\[t = \frac{5.0 \pm \sqrt{11.4}}{0.17}\]

04

Calculate the Valid Time

Compute the time values:\[t_1 = \frac{5.0 + 3.38}{0.17} \approx 48.07 \, s\]\[t_2 = \frac{5.0 - 3.38}{0.17} \approx 9.53 \, s\]The valid solution for catching up is \(t = 9.53 \,s\) since this represents the earliest point the student overtakes the bus.

05

Determine the Distance

Using the value of \(t = 9.53 \,s\), substitute back into the student's distance equation:\[d_s = 5.0 \cdot 9.53 \approx 47.65 \,m\]

06

Calculate Velocity of the Bus at Overtaking Point

The final velocity of the bus when the student overtakes it is given by:\[v = v_0 + at = 0 + 0.170 \cdot 9.53 \approx 1.62 \, \text{m/s}\]

07

Short Answer

(a) Time = 9.53 s, Distance = 47.65 m; (b) Velocity = 1.62 m/s. Further details and other parts require additional calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration

Understanding constant acceleration is crucial for solving problems involving objects that speed up or slow down at a predictable rate. Constant acceleration, often noted as "uniform acceleration," means that the rate of change of velocity over time remains consistent. This simplifies calculations because we can use the same formulas to predict future positions and velocities of moving objects.

In the exercise, the bus starts from rest and accelerates at a constant rate of 0.170 m/s². This implies its speed increases steadily every second by this amount. This is a typical scenario in kinematics: knowing the acceleration allows us to calculate the velocity and position of the bus at any given time.

• Key formulas involving constant acceleration include:
  • \( v = v_0 + at \) (Final velocity \( v \) given initial velocity \( v_0 \), acceleration \( a \), and time \( t \)).
  • \( d = v_0 t + \frac{1}{2} a t^2 \) (Distance \( d \) traveled given initial velocity, time, and acceleration).

These formulas give us the tools to determine how far an accelerating object, like the bus, has traveled or how fast it is going after a certain time.

Quadratic Equations in Physics

Many physics problems involving constant acceleration can lead to quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \), which can have at most two solutions. Understanding how to solve such equations is essential in kinematics, where motions are often not linear.

In the problem, the quadratic equation \( 0.085t^2 - 5.0t + 40.0 = 0 \) arises when setting the distances traveled by both the student and the bus equal to find when they overtake each other. Here, "\( a \)," "\( b \)," and "\( c \)" correspond to known values, and "\( t \)," the unknown, represents time.

• The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) helps solve for \( t \).
• Two values for \( t \) indicate when the student might pass the bus—once early on and possibly again at a later time if motion continues.

Problems with a second solution often involve scenarios where motion paths cross again, which might represent an opportunity for overtaking in repeated or circular tracks.

Learning to manipulate and apply these equations enhances problem-solving skills in physics and mathematics.

Relative Motion

Relative motion involves understanding how the position or velocity of one object is perceived from the position or velocity of another object. It’s key in scenarios where multiple objects are moving simultaneously.

In the exercise, the student and the bus are both in motion. The student's perspective of the bus changes as she sprints towards it. Initially, the student sees the bus as stationary with a starting distance of 40 meters.

• The critical aspect of solving the problem involves setting up equations where the relative distances of the student and the bus are equated.
• This allows us to determine the time at which they are at the same point, hence when overtaking occurs.

Understanding relative motion is essential in many real-life applications, such as determining when two moving objects in traffic might meet or estimating how quickly an athlete must run to reach a moving finish line on time.

Velocity Calculation

Velocity calculation determines how fast an object is moving in a particular direction. It is crucial for predicting future positions and assessing changes in motion.

For the student running towards the bus, her velocity is constant at 5.0 m/s. However, the bus's velocity is dynamic since it is accelerating. We calculate its velocity using the initial velocity, acceleration, and time. This is done using the formula:

• \( v = v_0 + at \)

At the moment the student catches the bus, substituting the time calculated earlier (9.53 seconds) into this equation gives the bus's velocity as \( 1.62 \, \text{m/s} \).

Understanding how to calculate velocity is not only important in kinematic problems but also useful in navigational systems, where knowing the speed and direction is fundamental for travel and logistics.