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Chapter 1: Problem 31

Average Velocity of a Motorcycle The distance \(s\) (in feet) covered by a motorcycle traveling in a straight line at any time \(t\) (in seconds) is given by the function $$ s(t)=-0.1 t^{3}+2 t^{2}+24 t $$ Calculate the motorcycle's average velocity over the time interval \([2,2+h]\) for \(h=1,0.1,0.01,0.001,0.0001\), and \(0.00001\), and use your results to guess at the motorcycle's instantaneous velocity at \(t=2\).

Short Answer

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#tag_title#Step 2: Calculate the average velocity \(v_{avg}\) #tag_content#Now we divide the displacement by the change in time to get the average velocity: $$v_{avg} = \frac{s(2 + h) - s(2)}{h}$$ #tag_title#Step 3: Plug in values of \(h\) and estimate instantaneous velocity at \(t = 2\) #tag_content#We calculate the average velocity for each value of \(h\): For \(h = 1\), $$v_{avg} = \frac{s(3) - s(2)}{1} \approx 30.5 \, \text{ft/s}$$ For \(h = 0.1\), $$v_{avg} = \frac{s(2.1) - s(2)}{0.1} \approx 41.59 \, \text{ft/s}$$ For \(h = 0.01\), $$v_{avg} = \frac{s(2.01) - s(2)}{0.01} \approx 43.8959 \, \text{ft/s}$$ For \(h = 0.001\), $$v_{avg} = \frac{s(2.001) - s(2)}{0.001} \approx 43.989959 \, \text{ft/s}$$ For \(h = 0.0001\), $$v_{avg} = \frac{s(2.0001) - s(2)}{0.0001} \approx 43.99899959 \, \text{ft/s}$$ For \(h = 0.00001\), $$v_{avg} = \frac{s(2.00001) - s(2)}{0.00001} \approx 43.9998999959 \, \text{ft/s}$$ As the value of \(h\) gets smaller, the average velocity approaches \(44 \, \text{ft/s}\). We can estimate the motorcycle's instantaneous velocity at \(t=2\) to be approximately \(44 \, \text{ft/s}\).

Step by step solution

01

Calculate the displacement 螖s

Using the given function, we calculate the displacement as follows: $$螖s = s(2+h) - s(2)$$

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Key Concepts

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

Instantaneous Velocity
To understand instantaneous velocity, think of it as the speed of an object at a particular moment in time. Unlike average velocity, which is calculated over a specific time interval, instantaneous velocity captures how fast something is moving right now. Imagine you're glancing at a car's speedometer; the value you see is an example of instantaneous velocity. You can't calculate it directly using basic arithmetic; that's where calculus steps in. Calculus allows us to find this value by computing the derivative of the function that describes the position over time. In our case about motorcycles, as we vary the value of \( h \) closer to zero, we determine the instantaneous velocity at \( t = 2 \). Here, we are essentially finding the slope of the tangent line to the curve at that point.
Displacement
Displacement refers to the change in position of the motorcycle along its path. It's important to note that displacement considers only the initial and final positions, ignoring the route taken. In mathematical terms, displacement is given by the difference \( 螖s = s(2+h) - s(2) \).

This formula involves calculating \( s(t) \) at two points. For our exercise, we illustrated how close the displacement comes to the instantaneous rate of change as \( h \) approaches zero. Displacement gives a broad view of movement over a particular interval; however, to find out how it changes at every instant, we look at derivatives.
Time Interval
The time interval dictates the duration over which certain measurements, like displacement and average velocity, are made. In our example with the motorcycle, the specified time interval ranges from \( [2, 2+h] \) for various values of \( h \). This range helps us gradually zoom in on the behavior of the velocity at \( t=2 \).

Understanding time intervals allows us to grasp how an average measurement becomes specific, transitioning to an instantaneous perspective. By shortening the time interval (decreasing \( h \) toward zero), we capture the behavior of the velocity with greater accuracy at a single point in time. This gives us clearer insights into the dynamics of movements like acceleration or deceleration at specific times.
Calculus
Calculus is the mathematical tool that helps us move from a discussion of averages to instantaneous values. It's like the microscope of mathematics, allowing us to examine changes at the tiniest scales. In our motorcycle scenario, calculus comes into play when we need to determine the instantaneous velocity, which requires taking the derivative of the position function \( s(t) \).

Derivatives in calculus represent how a function changes at any point and are crucial for finding rates like velocity. Here鈥檚 a simple way to think about it:
  • We calculate average rates over intervals, such as velocity over \([2, 2+h]\).
  • As \( h \) shrinks towards zero, the calculations become precise enough to reflect a single moment's rate, which is the instantaneous velocity.
Without calculus, these detailed insights into momentary changes would be much harder to obtain.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that shows it is false. The slope of the tangent line to the graph of \(f\) at the point \((a, f(a))\) is given by $$ \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} $$

In Exercises 57 and 58 , let \(f(x)=x\left(1-x^{2}\right)\), and let \(g\) be the signum (or sign) function defined by $$ g(x)=\left\\{\begin{array}{ll} -1 & \text { if } x<0 \\ 0 & \text { if } x=0 \\ 1 & \text { if } x>0 \end{array}\right. $$ Sketch the graph of the function \(g \circ f\), and determine where \(g \circ f\) is continuous.

Let $$ f(x)=\left\\{\begin{array}{ll} -1 & \text { if } x<0 \\ 1 & \text { if } x \geq 0 \end{array}\right. $$ Prove that \(\lim _{x \rightarrow 0} f(x)\) does not exist.

The position function of an object moving along a straight line is given by \(s=f(t) .\) The average velocity of the object over the time interval \([a, b]\) is the average rate of change of f over \([a, b] ;\) its (instantaneous) velocity at \(t=a\) is the rate of change of \(\bar{f}\) at \(a .\) During the construction of a high-rise building, a worker accidentally dropped his portable electric screwdriver from a height of \(400 \mathrm{ft}\). After \(t\) sec the screwdriver had fallen a distance of \(s=f(t)=16 t^{2} \mathrm{ft}\). a. How long did it take the screwdriver to reach the ground? b. What was the average velocity of the screwdriver during the time it was falling? c. What was the velocity of the screwdriver at the time it hit the ground?

Let $$ f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\ 1 & \text { if } x \text { is irrational } \end{array}\right. $$ Prove that \(\lim _{x \rightarrow 0} f(x)\) does not exist.
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