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Chapter 2: Problem 8

The function \(s(t)\) represents the position of an object at time \(t\) moving along a line. Suppose \(s(1)=84\) and \(s(4)=144 .\) Find the average velocity of the object over the interval of time \([1,4]\).

Short Answer

Expert verified
Answer: The average velocity of the object over the interval of time [1, 4] is 20 units per unit of time.

Step by step solution

01

Identify the positions at the given times

We are given \(s(1)=84\) and \(s(4)=144\). These are the positions of the object at times \(t=1\) and \(t=4\).
02

Calculate the difference in positions

To find the difference in positions, subtract the position at \(t=1\) from the position at \(t=4\): \(s(4)-s(1)=144-84=60\). So, the object has moved 60 units in the interval \([1,4]\).
03

Calculate the difference in time

The difference in time is the difference between the given times, which is \(4-1=3\). So, the interval of time is 3 units.
04

Calculate the average velocity

To calculate the average velocity, divide the difference in position by the difference in time: \(\frac{60}{3}=20\) units per unit of time. The average velocity of the object over the interval of time \([1,4]\) is 20 units per unit of time.

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Key Concepts

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

Understanding the Position Function
The concept of a position function, usually denoted as \(s(t)\), is used to describe the location of an object along a line at any given time \(t\). It provides a snapshot of where the object is positioned, enabling us to determine changes over specific intervals.
  • In the given exercise, the position function \(s(t)\) indicates the object's position at certain time points, specifically \(s(1)=84\) and \(s(4)=144\).
  • This means at time \(t = 1\), the object is at position 84, and at \(t = 4\), it is at position 144.
These snapshots allow you to track how far the object travels. By comparing positions at different times, we can calculate how much distance the object has covered over that interval.
Exploring the Interval of Time
The interval of time refers to the duration during which an event occurs or a change is observed. In this context, it represents the period between two time points that help us measure the change in position.
  • For the exercise at hand, the interval of time is given as \([1, 4]\), meaning from time \(t = 1\) to \(t = 4\).
  • This interval not only helps us identify the start and end time but also aids in calculating the time duration, which is a key input for further analysis.
To find the length of this interval, simply calculate the difference between these two time points: \(4 - 1 = 3\). This indicates that the change occurred over 3 time units.
Mastering Velocity Calculation
Calculating average velocity involves determining how quickly an object's position changes over a given time interval. The average velocity is found by taking the total change in position and dividing it by the total time taken.
  • In the exercise, we already calculated the change in position as \(s(4) - s(1) = 144 - 84 = 60\) units.
  • The time interval was determined to be 3 time units (from \(t = 1\) to \(t = 4\)).
With these values, the average velocity \(v_{avg}\) is calculated as \[v_{avg} = \frac{\text{change in position}}{\text{change in time}} = \frac{60}{3} = 20\text{ units per time unit}.\]Understanding this formula helps in settings where you have to figure out how fast or slow something moves over time. It is a foundational concept in physics and related fields where motion analysis is crucial.

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Most popular questions from this chapter

Suppose \(g(x)=\left\\{\begin{array}{ll}x^{2}-5 x & \text { if } x \leq-1 \\ a x^{3}-7 & \text { if } x>-1.\end{array}\right.\) Determine a value of the constant \(a\) for which \(\lim _{x \rightarrow-1} g(x)\) exists and state the value of the limit, if possible.

A monk set out from a monastery in the valley at dawn. He walked all day up a winding path, stopping for lunch and taking a nap along the way. At dusk, he arrived at a temple on the mountaintop. The next day the monk made the return walk to the valley, leaving the temple at dawn, walking the same path for the entire day, and arriving at the monastery in the evening. Must there be one point along the path that the monk occupied at the same time of day on both the ascent and descent? (Hint: The question can be answered without the Intermediate Value Theorem.) (Source: Arthur Koestler, The Act of Creation.)

a. Use the identity \(\sin (a+h)=\sin a \cos h+\cos a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \sin x=0\) to prove that \(\lim _{x \rightarrow a} \sin x=\sin a\) thereby establishing that \(\sin x\) is continuous for all \(x\). (Hint: Let \(h=x-a\) so that \(x=a+h\) and note that \(h \rightarrow 0\) as \(x \rightarrow a\).) b. Use the identity \(\cos (a+h)=\cos a \cos h-\sin a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \cos x=1\) to prove that \(\lim _{x \rightarrow a} \cos x=\cos a\).

Determine whether the following statements are true and give an explanation or counterexample. a. The value of \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) does not exist. b. The value of \(\lim _{x \rightarrow a} f(x)\) is always found by computing \(f(a)\) c. The value of \(\lim _{x \rightarrow a} f(x)\) does not exist if \(f(a)\) is undefined. d. \(\lim _{x \rightarrow 0} \sqrt{x}=0\) \(\lim _{x \rightarrow \pi / 2} \cot x=0\)

The Heaviside function is used in engineering applications to model flipping a switch. It is defined as $$H(x)=\left\\{\begin{array}{ll} 0 & \text { if } x<0 \\ 1 & \text { if } x \geq 0 \end{array}\right.$$ a. Sketch a graph of \(H\) on the interval [-1,1] b. Does \(\lim _{x \rightarrow 0} H(x)\) exist? Explain your reasoning after first examining \(\lim _{x \rightarrow 0^{-}} H(x)\) and \(\lim _{x \rightarrow 0^{+}} H(x)\)
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