Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 41
find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ \mathbf{r}(t)=3 t \mathbf{i}+3 t^{2} \mathbf{j} ; t_{1}=\frac{1}{3} $$
Short Answer
Expert verified
At \( t = \frac{1}{3} \), \( a_{T} = \frac{12}{\sqrt{13}} \) and \( a_{N} = \frac{6}{\sqrt{13}} \).
Step by step solution
01
Compute the Velocity Vector
To find the tangential and normal components, we first need the velocity vector. The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Compute \( \mathbf{v}(t) \):\[ \mathbf{v}(t) = \frac{d}{dt}(3t \mathbf{i} + 3t^2 \mathbf{j}) = 3 \mathbf{i} + 6t \mathbf{j}. \]
02
Compute the Acceleration Vector
Now, compute the acceleration vector \( \mathbf{a}(t) \). This is the derivative of the velocity vector \( \mathbf{v}(t) \). Find \( \mathbf{a}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(3 \mathbf{i} + 6t \mathbf{j}) = 6 \mathbf{j}. \]
03
Calculate the Tangential Component of Acceleration
The tangential component \( a_{T} \) is given by the projection of \( \mathbf{a}(t) \) onto \( \mathbf{v}(t) \). It can be calculated using the formula \( a_{T} = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \). First, compute \( \mathbf{a} \cdot \mathbf{v} \) and \( \|\mathbf{v}\| \): \[ \mathbf{a} \cdot \mathbf{v} = 0 \cdot 3 + 6 \cdot 6t = 36t. \]\[ \|\mathbf{v}\| = \sqrt{3^2 + (6t)^2} = \sqrt{9 + 36t^2}. \]So, \( a_{T} = \frac{36t}{\sqrt{9 + 36t^2}} \).
04
Calculate the Normal Component of Acceleration
The normal component \( a_{N} \) can be found using the formula \( a_{N} = \frac{\sqrt{\|\mathbf{a}\|^2 - (a_{T})^2}}{\|\mathbf{v}\|} \). First, compute \( \|\mathbf{a}\| \):\[ \|\mathbf{a}\| = \sqrt{0^2 + 6^2} = 6. \]Next, evaluate \( a_{N} = \frac{\sqrt{\|\mathbf{a}\|^2 - (a_{T})^2}}{\|\mathbf{v}\|} \) which simplifies to \( a_{N} = \frac{\sqrt{36 - \left( \frac{36t}{\sqrt{9 + 36t^2}} \right)^2}}{\sqrt{9 + 36t^2}} \).
05
Evaluate at t = t_1
Substitute \( t = \frac{1}{3} \) into the expressions for \( a_{T} \) and \( a_{N} \):For \( a_{T} \):\[ a_{T} = \frac{36 \times \frac{1}{3}}{\sqrt{9 + 36 \times \left(\frac{1}{3}\right)^2}} = \frac{12}{\sqrt{13}}. \]For \( a_{N} \): Plug \( t = \frac{1}{3} \) into the expression and simplify:\[ a_{N} = \frac{\sqrt{36 - \left( \frac{12}{\sqrt{13}} \right)^2}}{\sqrt{13}} = \frac{6\sqrt{1}}{\sqrt{13}} = \frac{6}{\sqrt{13}}. \]
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity Vector
The velocity vector is a core component in understanding motion. It represents how fast an object is moving and in which direction. When we talk about a velocity vector, we are looking at the rate of change of the position vector with respect to time. In mathematical terms, if we have a position vector \( \mathbf{r}(t) = 3t \mathbf{i} + 3t^2 \mathbf{j} \), the velocity vector \( \mathbf{v}(t) \) is the first derivative of this vector.
- Taking the derivative of \( \mathbf{r}(t) = 3t \mathbf{i} + 3t^2 \mathbf{j} \) gives \( \mathbf{v}(t) = 3 \mathbf{i} + 6t \mathbf{j} \).
- Here, \( 3 \mathbf{i} \) indicates a constant motion in the x-direction, while \( 6t \mathbf{j} \) shows that motion in the y-direction is speeding up over time.
Acceleration Vector
An acceleration vector indicates how the velocity of an object changes over time. It is essentially the derivative of the velocity vector. When we differentiate the velocity vector \( \mathbf{v}(t) = 3 \mathbf{i} + 6t \mathbf{j} \), we derive the acceleration vector \( \mathbf{a}(t) \).
- In our example, differentiating leads to \( \mathbf{a}(t) = 0 \mathbf{i} + 6 \mathbf{j} \).
- This shows there is no acceleration in the x-direction, while the acceleration in the y-direction is constant at \( 6 \mathbf{j} \).
Projection of Acceleration
The projection of acceleration is where we begin to break down the acceleration vector into components that line up with the motion trajectory. The tangential component \( a_T \), is essentially how much of the acceleration is acting in the same direction as the velocity vector.
- This is computed using the formula: \( a_T = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \).
- The dot product \( \mathbf{a} \cdot \mathbf{v} \) measures how aligned the acceleration vector is with the velocity vector.
- In many physics problems, the tangential acceleration gives insight into how the speed changes over time.
Vector Magnitude
In the analysis of dynamics, the concept of vector magnitude is crucial to understand the quantity, or "size", of vectors. For any vector \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \), its magnitude \( \|\mathbf{v}\| \) is calculated using the Pythagorean theorem: \( \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2} \).
- In the problem, the velocity vector's magnitude is \( \|\mathbf{v}\| = \sqrt{3^2 + (6t)^2} = \sqrt{9 + 36t^2} \).
- Magnitude helps determine the "speed" similar to how speed is the magnitude of velocity.
