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Chapter 3: Problem 56

Give the position \(s=f(t)\) of a body moving on a coordinate line ( \(s\) in meters, \(t\) in seconds). Find the body's velocity, speed, acceleration, and jerk at time \(t=\pi / 4 \mathrm{s}\). $$s=\sin t+\cos t$$

Short Answer

Expert verified
Velocity and jerk are 0, speed is 0, acceleration is \(-\sqrt{2}\) at \(t=\frac{\pi}{4}\).

Step by step solution

01

Find the Velocity

The velocity is the first derivative of the position function with respect to time. Given \(s(t) = \sin t + \cos t\), we can differentiate with respect to \(t\) to find:\[ v(t) = \frac{d}{dt}(\sin t + \cos t) = \cos t - \sin t. \]Substitute \(t = \frac{\pi}{4}\) into the velocity expression:\[ v\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right). \]Since \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\) and \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we have:\[ v\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0. \]
02

Calculate the Speed

Speed is the absolute value of velocity. From Step 1, the velocity at \(t = \frac{\pi}{4}\) is \(0\). Therefore, the speed is:\[ |v\left(\frac{\pi}{4}\right)| = |0| = 0. \]
03

Determine the Acceleration

Acceleration is the derivative of velocity with respect to time. Starting with \(v(t) = \cos t - \sin t\), differentiate to get acceleration:\[ a(t) = \frac{d}{dt}(\cos t - \sin t) = -\sin t - \cos t. \]Evaluate this at \(t = \frac{\pi}{4}\):\[ a\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}. \]
04

Compute the Jerk

Jerk is the derivative of the acceleration with respect to time. From \(a(t) = -\sin t - \cos t\), differentiate to find jerk:\[ j(t) = \frac{d}{dt}(-\sin t - \cos t) = -\cos t + \sin t. \]Evaluate this at \(t = \frac{\pi}{4}\):\[ j\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0. \]

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Key Concepts

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

Understanding Velocity
Velocity is a cornerstone concept in calculus of motion. It represents how fast and in what direction a body is moving. Mathematically, velocity is the first derivative of the position function with respect to time. In our exercise, the position function is given by: \[ s(t) = \sin t + \cos t \] By taking the first derivative with respect to time \( t \), you get the velocity function: \[ v(t) = \frac{d}{dt}(\sin t + \cos t) = \cos t - \sin t \] Evaluating this at \( t = \frac{\pi}{4} \), we substitute \( \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \) and \( \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \), which gives: - \( v \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \)
  • The zero result means the body is stationary at this instant.
  • Speed is the magnitude of velocity. Here, speed = \( |0| = 0 \).
Exploring Acceleration
Acceleration quantifies changes in velocity over time. It's the derivative of velocity concerning time. Given our velocity function from before, \[ v(t) = \cos t - \sin t \], differentiate to find the acceleration: \[ a(t) = \frac{d}{dt}(\cos t - \sin t) = -\sin t - \cos t \] Evaluate at \( t = \frac{\pi}{4} \): \[ a \left( \frac{\pi}{4} \right) = -\sin \left( \frac{\pi}{4} \right) - \cos \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \]
Acceleration tells us about the increase or decrease in speed. A negative value indicates the body slows down or speeds up in the opposite direction.
Understanding Jerk
Jerk represents the rate of change of acceleration. It's the third derivative, showing how acceleration is evolving over time. With acceleration being \[ a(t) = -\sin t - \cos t \], differentiate to find the jerk: \[ j(t) = \frac{d}{dt}(-\sin t - \cos t) = -\cos t + \sin t \] Substitute \( t = \frac{\pi}{4} \) to evaluate the jerk: \[ j \left( \frac{\pi}{4} \right) = -\cos \left( \frac{\pi}{4} \right) + \sin \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \]
  • A jerk of zero indicates that the acceleration remains constant at that instant.
  • Understanding these derivatives helps in analyzing motion precisely, as it depicts not just the current state but also how it is changing.

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Most popular questions from this chapter

Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

Graph \(y=(\sin x) / x, y=(\sin 2 x) / x,\) and \(y=(\sin 4 x) / x\) together over the interval \(-2 \leq x \leq 2 .\) Where does each graph appear to cross the \(y\) -axis? Do the graphs really intersect the axis? What would you expect the graphs of \(y=(\sin 5 x) / x\) and \(y=(\sin (-3 x)) / x\) to do as \(x \rightarrow 0 ?\) Why? What about the graph of \(y=(\sin k x) / x\) for other values of \(k ?\) Give reasons for your answers.

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 \(x=3 \mathrm{m} ?\)

Graph \(y=\tan x\) and its derivative together on \((-\pi / 2, \pi / 2) .\) Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.

The coordinates of a particle in the metric \(x y\) -plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{s}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{s} .\) How fast is the particle's distance from the origin changing as it passes through the point (5,12)\(?\)
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