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Chapter 3: Problem 6

give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$s=\frac{25}{t+5}, \quad-4 \leq t \leq 0$$

Short Answer

Expert verified
a. Displacement: -20 m, Avg. velocity: -5 m/s. b. Speeds: 25 m/s, 1 m/s; Acceleration: 50, 0.4. c. No direction change.

Step by step solution

01

Define the Displacement Function

The displacement of a body on a coordinate line is defined by the function \( s = \frac{25}{t+5} \). The displacement at the endpoint \( t=-4 \) is \( s(-4) = \frac{25}{-4+5} = \frac{25}{1} = 25 \). Similarly, the displacement at \( t=0 \) is \( s(0) = \frac{25}{0+5} = \frac{25}{5} = 5 \).
02

Calculate the Displacement

The body's displacement over the interval \([-4, 0]\) seconds is the difference in position at the endpoints: \(s_2 - s_1 = s(0) - s(-4) = 5 - 25 = -20 \text{ meters}\). This indicates that the body has moved 20 meters backwards.
03

Find the Average Velocity

Average velocity is the total displacement divided by the time interval. Here, the displacement \(\Delta s = -20\) meters and the time interval \(\Delta t = 0 - (-4) = 4\) seconds. So, the average velocity is \[ v_{avg} = \frac{\Delta s}{\Delta t} = \frac{-20}{4} = -5 \text{ m/s}\]. This negative value indicates movement in the negative direction.
04

Compute Speed and Acceleration at \(t=-4\) and \(t=0\)

The speed is the absolute value of velocity, so we first need the velocity function: \( v(t) = \frac{d}{dt}\left(\frac{25}{t+5}\right) = -\frac{25}{(t+5)^2} \). At \(t=-4\), \(v(-4) = -\frac{25}{1^2} = -25 \text{ m/s}\), and at \(t=0\), \(v(0) = -\frac{25}{5^2} = -1 \text{ m/s}\). Thus, speed at \(t=-4\) is 25 m/s and at \(t=0\) is 1 m/s. To find acceleration, compute the derivative of velocity: \[ a(t) = \frac{d}{dt}\left(-\frac{25}{(t+5)^2}\right) = \frac{50}{(t+5)^3} \]. At \(t=-4\), \(a(-4) = 50\) and at \(t=0\), \(a(0) = 0.4\).
05

Determine Direction Change

The body changes direction when the velocity is zero within the interval \([-4, 0]\). Solve \(-\frac{25}{(t+5)^2} = 0\), but this has no real solutions. Thus, the body never changes direction because the velocity does not become zero within the given time interval.

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Key Concepts

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

Understanding Displacement in Kinematics
Displacement is a fundamental concept in kinematics and describes the change in position of a body over a specific time interval. It is a vector quantity, meaning it has both magnitude and direction. In the exercise, the displacement function is given by \( s = \frac{25}{t+5} \) for \( -4 \leq t \leq 0 \). To find displacement, we need to evaluate the position of the body at both the start and the end of this interval.
For \( t = -4 \), the position \( s(-4) \) is calculated to be 25 meters and for \( t = 0 \), \( s(0) \) is 5 meters. Therefore, displacement \( \Delta s \) is the difference between these positions:
\[ \Delta s = s(0) - s(-4) = 5 - 25 = -20 \text{ meters} \].
This result indicates that the body has moved 20 meters in the negative direction during the interval.
Calculating Average Velocity
Average velocity is crucial in understanding how fast something moves over an interval, regardless of specific momentary changes in speed. It is defined as the total displacement divided by the total time taken. In our exercise, we already found that the displacement is \(-20 \text{ meters}\) over a \(4\text{-second}\) interval.
The formula for average velocity \( v_{avg} \) is:
\[ v_{avg} = \frac{\Delta s}{\Delta t} = \frac{-20}{4} = -5 \text{ m/s} \].
This means, on average, the body moves five meters per second in the negative direction. The negative sign indicates the direction of the velocity aligns with the direction of the displacement.
Understanding Acceleration in Motion
Acceleration refers to how quickly an object changes its velocity. It is a vector quantity, meaning it also has direction. To compute acceleration, we take the derivative of velocity with respect to time. From the velocity function \( v(t) = -\frac{25}{(t+5)^2} \), we derived the acceleration function:
\[ a(t) = \frac{50}{(t+5)^3} \].
For the given times, the acceleration was calculated as follows:
  • At \( t = -4\), \(a(-4) = 50\) m/s虏.
  • At \( t = 0\), \(a(0) = 0.4\) m/s虏.

This shows that the acceleration is much higher at the beginning of the interval, indicating the body is initially decelerating rapidly. As the body progresses to \( t = 0 \), the acceleration decreases, pointing towards a slowing change in velocity.

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Most popular questions from this chapter

For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L\). In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2\)

Find \(d y / d t\) $$y=4 \sin (\sqrt{1+\sqrt{t}})$$

A particle moves along the parabola \(y=x^{2}\) in the first quadrant in such a way that its \(x\) -coordinate (measured in meters) increases at a steady \(10 \mathrm{m} / \mathrm{s}\). How fast is the angle of inclination \(\theta\) of the line joining the particle to the origin changing when \(x=3 \mathrm{m} ?\)

a. Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with the properties: i) \(Q(a)=f(a)\) ii) \(Q^{\prime}(a)=f^{\prime}(a)\) iii) \(Q^{\prime \prime}(a)=f^{\prime \prime}(a)\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) b. Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0\) c. Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point (0,1) Comment on what you see. d. Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) Graph \(g\) and its quadratic approximation together. Comment on what you see. e. Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. f. What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts (b), (d), and (e)?

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