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Chapter 4: Problem 39

Frictionless Cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time \(t=0\) to roll back and forth for 4 sec. Its position at time \(t\) is \(s=10 \cos \pi t .\) (a) What is the cart's maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? (b) Where is the cart when the magnitude of the acceleration is greatest? What is the cart's speed then?

Short Answer

Expert verified
The cart's maximum speed is \(10 \pi\) cm/s and it reaches this speed at \(t = 1/2\) and \(t = 3/2\) sec. At those times it's at equilibrium position and the acceleration is \(10 \pi^2\) cm/s虏. The acceleration is greatest, \(10 \pi^2\) cm/s虏, when the cart is at its extreme positions, 10 cm from equilibrium, where its speed is zero.

Step by step solution

01

Derive the Function of Velocity

To find the maximum speed, think of speed as the magnitude of the velocity. We can derive velocity as the derivative of the position function. Let's differentiate the position function \(s=10 \cos \pi t\), we get \(v = -10 \pi \sin \pi t\).
02

Find the Maximum Speed and the Time When it's Reached

The magnitude of the velocity gives us the speed. For any \(t\), the maximum value of \(\sin \pi t\) (and hence of \(\sin \pi t\)) is 1, so the maximum speed is \(10 \pi\) cm/s. The \(\sin \pi t\) obtains its maximum value when its argument \(\pi t\) equals \( \frac{\pi}{2}\), or \( \frac{3\pi}{2}\), 't' can be either \(\frac{1}{2}\) or \(\frac{3}{2}\) sec for a 4-sec interval.
03

Find the Position of Cart and the Magnitude of Acceleration at Maximum Speed

To find where the cart is at maximum speed, we substitute \(t = \frac{1}{2}\) and \(t = \frac{3}{2}\) into our position function. We find both times give \(s = 0\). To find the acceleration when cart is at maximum speed, we need to derive the velocity function, that gives us \(a = -(10 \pi^2) \cos \pi t\). Substituting \(t = \frac{1}{2}\) and \(t = \frac{3}{2}\) into our acceleration function, we find both times give \(a = 10 \pi^2\) cm/s虏.
04

Find the Position and Speed When Acceleration is Maximum

The magnitude of the acceleration reaches maximum value when value of \(\cos \pi t\) is maximum, i.e. 1. This happens when \(t = 0\), or \(2\), so at both ends of the 4-sec interval. In both cases, the cart is at its extreme position, 10 cm from the equilibrium position. The cart's speed at those times is zero because it's momentarily at rest.

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Key Concepts

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

Derivatives in Physics
In physics, derivatives help us understand how things change over time. When we deal with motion, the derivative of the position function gives us the velocity. It's similar to how you would describe speed in everyday terms as how quickly you鈥檙e moving.

Let's break it down:
  • The position function describes where an object is at any given time.
  • By taking the derivative, you're calculating the velocity, or how fast the position is changing.
In this exercise, the derivative of the cart's position 鈥� given by the equation \(s = 10 \cos \pi t\) 鈥� helps us find velocity: \(v = -10 \pi \sin \pi t\).

So, if you're trying to find how rapidly the cart is moving, derivatives provide the answer. It's essentially a snapshot of motion at any point in time, making it easier to see how fast or slow that motion is happening.
Maximum Speed
Finding an object's maximum speed requires understanding when its velocity reaches its peak value. When we say maximum speed, we refer to the highest speed obtainable, irrespective of direction.

For the given scenario, you calculate speed by finding the magnitude of the velocity function we've derived: \(v = -10 \pi \sin \pi t\).

Consider these points:
  • The function \(\sin \pi t\) oscillates between 0 and 1.
  • The maximum speed occurs when \(\sin \pi t\) equals 1, making \(v\) reach \(10 \pi\) cm/s.
Therefore, the maximum speed of the cart is \(10 \pi\) cm/s. Understanding when the trigonometric function hits its peak allows us to pinpoint exact timings like \(t = \frac{1}{2}\) or \(t = \frac{3}{2}\) seconds. At these times, the cart moves fastest.
Acceleration
Acceleration tells us how quickly the speed of an object is changing. In other words, it's the derivative of the velocity function. So, to find out how the cart's speed changes, we derive our velocity equation again.

Starting with \(v = -10 \pi \sin \pi t\), the derivative gives us the acceleration: \(a = -(10 \pi^2) \cos \pi t\). This equation shows how the cart speeds up or slows down over time.

Here鈥檚 a breakdown:
  • The acceleration magnitude reaches its highest when \(\cos \pi t\) is maximum, which is 1.
  • At these points, like \(t = 0\) or \(2\), the acceleration is maximized. The cart is momentarily at rest, yet acceleration is at its peak because it's about to change motion direction.
Thus, acceleration provides an excellent insight into the dynamics of oscillatory movements.
Trigonometric Functions
In oscillatory motion, trigonometric functions play a crucial role. They describe periodic movements such as the direction and distance of oscillation.

For the problem at hand, functions like \(\cos \pi t\) and \(\sin \pi t\) help simulate the cart's back-and-forth motion:

  • \(\cos \pi t\) indicates the cart's position concerning time, ensuring it swings between maximum displacements.
  • \(\sin \pi t\)'s peak values determine when the cart reaches its fastest speed.

Understanding these harmonious functions offers insight into predicting motion behaviors without manual calculation at every step. These periodic functions model real-world oscillations efficiently, from simple pendulums to electrical waves, underscoring their immense utility in physics.

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Most popular questions from this chapter

Measuring Acceleration of Gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\) . The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g\) . By keeping track of \(\Delta T\) , we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) (a) With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts \((b)\) and \((c)\) . (b) Writing to Learn If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. (c) A clock with a 100 -cm pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.

True or False If \(f^{\prime}(c)=0\) and \(f^{\prime \prime}(c)<0,\) then \(f(c)\) is a local maximum. Justify your answer.

Minting Coins A manufacturer contracts to mint coins for the federal government. The coins must weigh within 0.1\(\%\) of their ideal weight, so the volume must be within 0.1\(\%\) of the ideal volume. Assuming the thickness of the coins does not change, what is the percentage change in the volume of the coin that would result from a 0.1\(\%\) increase in the radius?

Writing to Learn You have been asked to determine whether the function \(f(x)=3+4 \cos x+\cos 2 x\) is ever negative. (a) Explain why you need to consider values of \(x\) only in the interval \([0,2 \pi] . \quad\) (b) Is f ever negative? Explain.

$$ \begin{array}{l}{\text { Analyzing Motion Data Priya's distance } D \text { in meters from a }} \\ {\text { motion detector is given by the data in Table 4.1. }}\end{array} $$ $$ \begin{array}{llll}{t(\text { sec) }} & {D(\mathrm{m})} & {t(\mathrm{sec})} & {D(\mathrm{m})} \\ \hline 0.0 & {3.36} & {4.5} & {3.59} \\ {0.5} & {2.61} & {5.0} & {4.15} \\ {1.0} & {1.86} & {5.5} & {3.99} \\ {1.5} & {1.27} & {6.0} & {3.37}\end{array} $$ $$ \begin{array}{llll}{2.0} & {0.91} & {6.5} & {2.58} \\ {2.5} & {1.14} & {7.0} & {1.93} \\ {3.0} & {1.69} & {7.5} & {1.25} \\ {3.5} & {2.37} & {8.0} & {0.67} \\ {4.0} & {3.01}\end{array} $$ $$ \begin{array}{l}{\text { (a) Estimate when Priya is moving toward the motion detector; }} \\ {\text { away from the motion detector. }} \\ {\text { (b) Writing to Learn Give an interpretation of any local }} \\ {\text { extreme values in terms of this problem situation. }}\end{array} $$ $$ \begin{array}{l}{\text { (c) Find a cubic regression equation } D=f(t) \text { for the data in }} \\ {\text { Table } 4.1 \text { and superimpose its graph on a scatter plot of the data. }} \\ {\text { (d) Use the model in (c) for } f \text { to find a formula for } f^{\prime} . \text { Use this }} \\ {\text { formula to estimate the answers to (a). }}\end{array} $$
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