/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Velocity and acceleration: diffe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Velocity and acceleration: differentiation of vectors. Find the velocity and acceleration of the point described by the following position vectors \((t=\) time in seconds); (a) \(\mathbf{r}=16 t \hat{\mathbf{x}}+25 t^{2} \hat{\mathbf{y}}+33 \hat{\mathbf{z}} \mathrm{cm}\) (b) \(\quad \mathbf{r}=10 \sin 15 t \hat{\mathbf{x}}+35 t \hat{\mathbf{y}}+e^{6 i \hat{\mathbf{z}}} \mathrm{cm}\) (For derivatives see the Mathematical Notes at the end of this chapter.)

Short Answer

Expert verified
Velocity (a): \(16 \hat{\mathbf{x}} + 50t \hat{\mathbf{y}}\), Acceleration (a): \(50 \hat{\mathbf{y}}\). Velocity (b): \(150 \cos(15t) \hat{\mathbf{x}} + 35 \hat{\mathbf{y}} + 6e^{6t} \hat{\mathbf{z}}\), Acceleration (b): \(-2250 \sin(15t) \hat{\mathbf{x}} + 36e^{6t} \hat{\mathbf{z}}\).

Step by step solution

01

Understand the Problem

We need to find the velocity and acceleration vectors for a particle described by its position vector as functions of time. The velocity is the first derivative of the position vector with respect to time, and the acceleration is the second derivative of the position vector with respect to time.
02

Differentiate Position Vector (a) to Find Velocity

For position vector (a), \[ \mathbf{r} = 16 t \hat{\mathbf{x}} + 25 t^2 \hat{\mathbf{y}} + 33 \hat{\mathbf{z}} \]Differentiate each component with respect to time:- \(\frac{d}{dt}(16t) = 16\)- \(\frac{d}{dt}(25t^2) = 50t\)- \(\frac{d}{dt}(33) = 0\)Thus, the velocity vector is:\[ \mathbf{v} = 16 \hat{\mathbf{x}} + 50t \hat{\mathbf{y}} + 0 \hat{\mathbf{z}} \]
03

Differentiate Velocity Vector (a) to Find Acceleration

Differentiate the velocity vector \(\mathbf{v} = 16 \hat{\mathbf{x}} + 50t \hat{\mathbf{y}} + 0 \hat{\mathbf{z}}\):- \(\frac{d}{dt}(16) = 0\)- \(\frac{d}{dt}(50t) = 50\)- \(\frac{d}{dt}(0) = 0\)Thus, the acceleration vector is:\[ \mathbf{a} = 0 \hat{\mathbf{x}} + 50 \hat{\mathbf{y}} + 0 \hat{\mathbf{z}} \]
04

Differentiate Position Vector (b) to Find Velocity

For position vector (b),\[ \mathbf{r} = 10 \sin(15t) \hat{\mathbf{x}} + 35t \hat{\mathbf{y}} + e^{6t} \hat{\mathbf{z}} \]Differentiate each component with respect to time:- \(\frac{d}{dt}(10 \sin(15t)) = 150 \cos(15t)\)- \(\frac{d}{dt}(35t) = 35\)- \(\frac{d}{dt}(e^{6t}) = 6e^{6t}\)Thus, the velocity vector is:\[ \mathbf{v} = 150 \cos(15t) \hat{\mathbf{x}} + 35 \hat{\mathbf{y}} + 6e^{6t} \hat{\mathbf{z}} \]
05

Differentiate Velocity Vector (b) to Find Acceleration

Differentiate the velocity vector \(\mathbf{v} = 150 \cos(15t) \hat{\mathbf{x}} + 35 \hat{\mathbf{y}} + 6e^{6t} \hat{\mathbf{z}}\):- \(\frac{d}{dt}(150 \cos(15t)) = -2250 \sin(15t)\)- \(\frac{d}{dt}(35) = 0\)- \(\frac{d}{dt}(6e^{6t}) = 36e^{6t}\)Thus, the acceleration vector is:\[ \mathbf{a} = -2250 \sin(15t) \hat{\mathbf{x}} + 0 \hat{\mathbf{y}} + 36e^{6t} \hat{\mathbf{z}} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Velocity and Acceleration
In kinematics, velocity and acceleration are key concepts used to describe the motion of objects. Velocity is the rate at which an object's position changes over time, while acceleration is the rate at which its velocity changes.
To determine the velocity of an object, we find the first derivative of the position vector with respect to time. This involves differentiating each component of the position vector, such as (a) \(\mathbf{r} = 16t \hat{\mathbf{x}} + 25t^2 \hat{\mathbf{y}} + 33 \hat{\mathbf{z}}\), where the derivatives are \(16 \hat{\mathbf{x}}, 50t \hat{\mathbf{y}},\) and \(0 \hat{\mathbf{z}}\).
The acceleration is found by differentiating the velocity vector. From our earlier example, the velocity is \(\mathbf{v} = 16 \hat{\mathbf{x}} + 50t \hat{\mathbf{y}} + 0 \hat{\mathbf{z}}\). Its derivative gives us the acceleration \(\mathbf{a} = 0 \hat{\mathbf{x}} + 50 \hat{\mathbf{y}} + 0 \hat{\mathbf{z}}\), which clarifies the change in velocity over time. Understanding these concepts allows us to predict the future motion of an object.
Vector Differentiation
Vector differentiation is essential when dealing with motion in multiple dimensions, such as in physics and engineering. A vector consists of components along different axes, for instance, \(\hat{\mathbf{x}}, \hat{\mathbf{y}},\) and \(\hat{\mathbf{z}}\).
To differentiate a vector, differentiate each component independently. For example, take the vector \(\mathbf{r} = 10 \sin(15t) \hat{\mathbf{x}} + 35t \hat{\mathbf{y}} + e^{6t} \hat{\mathbf{z}}\):
  • For \(10 \sin(15t) \hat{\mathbf{x}}\), the derivative is \(150 \cos(15t) \hat{\mathbf{x}}\).
  • For \(35t \hat{\mathbf{y}}\), it is constant and its derivative is \(35 \hat{\mathbf{y}}\).
  • The exponential component \(e^{6t} \hat{\mathbf{z}}\) becomes \(6e^{6t} \hat{\mathbf{z}}\).
This gives us the velocity vector, which describes the object's motion in three dimensions at any given time.
Derivatives in Physics
Derivatives are a mathematical tool that helps us understand how quantities change over time, particularly in physics. They play a critical role in analyzing motion, encapsulating concepts like velocity, acceleration, and more.
In the context of kinematics, the first derivative of a position function gives velocity, indicating how fast and in what direction something moves. Taking the second derivative results in acceleration, a measure of how rapidly the velocity is changing.
Consider position function \(\mathrm{r}(t)\), differentiating gives velocity \(\frac{dr}{dt}\), and further differentiating gives acceleration \(\frac{d^2r}{dt^2}\). Knowing these derivatives allows scientists and engineers to design better systems and predict motions accurately, from the path of a projectile to the orbit of a satellite.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Scalar and vector products of two vectors. Given two vectors \(\mathbf{a}=3 \hat{\mathbf{x}}+4 \hat{y}-5 \hat{\mathbf{z}}\) and \(\mathbf{b}=-\hat{\mathbf{x}}+2 \hat{\mathbf{y}}+6 \hat{\mathbf{z}}\), calculate by vector methods: (a) The length of each Ans, \(a=\sqrt{50} ; b=\sqrt{41}\). (b) The scalar product a \cdot b Ans. \(-25\). (c) The included angle between them Ans. \(123.5^{\circ}\). (d) The direction cosines for each (e) The vector sum and difference \(a+b\) and \(a-b\)

Position cectors. Using the \(x\) axis as east, the \(y\) axis as north, and the \(z\) axis as up, give the vector representing the following points: (a) \(10 \mathrm{mi}\) northeast and \(2 \mathrm{mi}\) up (b) \(5 \mathrm{yd}\) southeast and \(5 \mathrm{yd}\) down (c) \(1 \mathrm{~cm}\) northwest and \(6 \mathrm{~cm}\) up Find the magnitude of each vector and the expression for the unit vector in that direction.

Random flights. A particle follows in space a path that consists of \(N\) equal steps, each of length \(s .\) The direction in space of each step is entirely random, with no relation or correlation between any two steps. The total displacement is $$ \mathbf{S}=\sum_{i=1}^{N} \mathbf{s}_{i} $$ Show that the mean square displacement between initial and final positions is \(\left\langle\mathrm{S}^{2}\right\rangle=\mathrm{Ns}^{2}\), where \langle\rangle denotes mean value. [Hint: The assumption that the direction of every step is independent of the direction of every other step means that \(\left\langle s_{i} \cdot \mathrm{s}_{j}\right\rangle=0\) for all \(i\) and \(i\), except \(\left.i=j .\right]\)

Multiplication by a scalar. Let \(\mathbf{A}=2.0 \mathrm{~cm}\) at \(70^{\circ}\) east of north and \(B=3.5 \mathrm{~cm}\) at \(130^{\circ}\) east of north. Use either a protractor or polar coordinate graph paper in your solutions. (a) Draw the vectors described above and two others \(2.5\) times as large. (b) Multiply A by \(-2\) and \(\mathbf{B}\) by \(+3\) and find the vector sum. Ans. \(9.2 \mathrm{~cm}\) at \(152^{\circ}\). (c) Place a point \(10 \mathrm{~cm}\) due north of the origin. Find multiples of \(\mathrm{A}\) and \(\mathrm{B}\) whose vector sum is the vector from the origin to this point. (d) Work out parts \((b)\) and \((c\rangle\) analytically.

Addition of cectors. Draw the result of the following vector additions: (a) Add a vector \(2 \mathrm{~cm}\) east to one \(3 \mathrm{~cm}\) northwest. (b) Add a vector \(8 \mathrm{~cm}\) east to one \(12 \mathrm{~cm}\) northwest. (c) Compare the results of parts \((a)\) and \((b)\), and frame a theorem about adding a pair of vectors that are multiples of another pair.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.