The equation r(t) = (t) i + (V3t- 16t) j is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t = 0. The angle between the velocity and acceleration vectors at time t= 0 is radians. (Type an exact answer, using n as needed.)
The equation r(t) = (t) i + (V3t- 16t) j is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t = 0. The angle between the velocity and acceleration vectors at time t= 0 is radians. (Type an exact answer, using n as needed.)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Problem Statement:
Consider a particle moving in space whose position at any given time \( t \) is given by the equation:
\[ \mathbf{r}(t) = t \mathbf{i} + \left( \sqrt{3}t - 16t^2 \right) \mathbf{j} \]
where \( \mathbf{i} \) and \( \mathbf{j} \) are the unit vectors in the x and y directions, respectively.
You are required to find the angle between the velocity and acceleration vectors at \( t = 0 \).
### Solution:
#### Step 1: Determine the Velocity Vector
The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \):
\[ \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \]
\[ \mathbf{v}(t) = \frac{d}{dt} \left( t \mathbf{i} + \left( \sqrt{3}t - 16t^2 \right) \mathbf{j} \right) \]
\[ \mathbf{v}(t) = \mathbf{i} + \left( \sqrt{3} - 32t \right) \mathbf{j} \]
#### Step 2: Determine the Acceleration Vector
The acceleration vector \( \mathbf{a}(t) \) is the first derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \):
\[ \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \]
\[ \mathbf{a}(t) = \frac{d}{dt} \left( \mathbf{i} + \left( \sqrt{3} - 32t \right) \mathbf{j} \right) \]
\[ \mathbf{a}(t) = -32 \mathbf{j} \]
#### Step 3: Evaluate the Vectors at \( t = 0 \)
At \( t = 0 \):
\[ \mathbf{v}(0) = \mathbf{i} + \left( \sqrt{3} - 32(0) \right) \mathbf{j} = \mathbf{i} + \sqrt{3} \mathbf{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9d15e77-8c44-4e48-af9a-bae33a9e346c%2Fcba71f99-ab5a-4d39-b5d0-b587b16d45fb%2Frzunyh8_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Consider a particle moving in space whose position at any given time \( t \) is given by the equation:
\[ \mathbf{r}(t) = t \mathbf{i} + \left( \sqrt{3}t - 16t^2 \right) \mathbf{j} \]
where \( \mathbf{i} \) and \( \mathbf{j} \) are the unit vectors in the x and y directions, respectively.
You are required to find the angle between the velocity and acceleration vectors at \( t = 0 \).
### Solution:
#### Step 1: Determine the Velocity Vector
The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \):
\[ \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \]
\[ \mathbf{v}(t) = \frac{d}{dt} \left( t \mathbf{i} + \left( \sqrt{3}t - 16t^2 \right) \mathbf{j} \right) \]
\[ \mathbf{v}(t) = \mathbf{i} + \left( \sqrt{3} - 32t \right) \mathbf{j} \]
#### Step 2: Determine the Acceleration Vector
The acceleration vector \( \mathbf{a}(t) \) is the first derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \):
\[ \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \]
\[ \mathbf{a}(t) = \frac{d}{dt} \left( \mathbf{i} + \left( \sqrt{3} - 32t \right) \mathbf{j} \right) \]
\[ \mathbf{a}(t) = -32 \mathbf{j} \]
#### Step 3: Evaluate the Vectors at \( t = 0 \)
At \( t = 0 \):
\[ \mathbf{v}(0) = \mathbf{i} + \left( \sqrt{3} - 32(0) \right) \mathbf{j} = \mathbf{i} + \sqrt{3} \mathbf{
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