At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Certainly! Let's solve this step-by-step based on the given problem.
### Problem Statement
The motion of a particle in a plane is defined by the equations:
[tex]$ x(t) = 13 + 3t - 4 $[/tex]
[tex]$ y(t) = 12 - 2t + 6 $[/tex]
We need to determine the velocity and acceleration of the particle after 3 seconds (t = 3 seconds).
### Step 1: Calculate the Position at t = 3 seconds
We need to determine the position of the particle at [tex]\( t = 3 \)[/tex] seconds using the given equations for [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex].
#### For x(t):
[tex]$ x(3) = 13 + 3(3) - 4 $[/tex]
[tex]\[ = 13 + 9 - 4 \\ = 18 \][/tex]
#### For y(t):
[tex]$ y(3) = 12 - 2(3) + 6 $[/tex]
[tex]\[ = 12 - 6 + 6 \\ = 12 \][/tex]
So, the position of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (18, 12) \)[/tex] meters.
### Step 2: Determine the Velocity
Velocity is the first derivative of the position with respect to time.
#### For x(t):
The velocity in the x-direction, [tex]\( v_x \)[/tex], is the derivative of [tex]\( x(t) \)[/tex]:
[tex]$ \frac{dx}{dt} = \frac{d}{dt}(13 + 3t - 4) \\ = 3 $[/tex]
#### For y(t):
The velocity in the y-direction, [tex]\( v_y \)[/tex], is the derivative of [tex]\( y(t) \)[/tex]:
[tex]$ \frac{dy}{dt} = \frac{d}{dt}(12 - 2t + 6) \\ = -2 $[/tex]
Therefore, the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (3, -2) \)[/tex] meters per second.
### Step 3: Determine the Acceleration
Acceleration is the second derivative of the position with respect to time or the first derivative of velocity with respect to time.
#### For x(t):
The acceleration in the x-direction, [tex]\( a_x \)[/tex], is the derivative of [tex]\( v_x \)[/tex]:
[tex]$ \frac{d^2x}{dt^2} = \frac{d}{dt}(3) \\ = 0 $[/tex]
#### For y(t):
The acceleration in the y-direction, [tex]\( a_y \)[/tex], is the derivative of [tex]\( v_y \)[/tex]:
[tex]$ \frac{d^2y}{dt^2} = \frac{d}{dt}(-2) \\ = 0 $[/tex]
Therefore, the acceleration of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (0, 0) \)[/tex] meters per second squared.
### Summary
- Position at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (18, 12) \)[/tex] meters
- Velocity at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (3, -2) \)[/tex] meters per second
- Acceleration at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (0, 0) \)[/tex] meters per second squared
### Problem Statement
The motion of a particle in a plane is defined by the equations:
[tex]$ x(t) = 13 + 3t - 4 $[/tex]
[tex]$ y(t) = 12 - 2t + 6 $[/tex]
We need to determine the velocity and acceleration of the particle after 3 seconds (t = 3 seconds).
### Step 1: Calculate the Position at t = 3 seconds
We need to determine the position of the particle at [tex]\( t = 3 \)[/tex] seconds using the given equations for [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex].
#### For x(t):
[tex]$ x(3) = 13 + 3(3) - 4 $[/tex]
[tex]\[ = 13 + 9 - 4 \\ = 18 \][/tex]
#### For y(t):
[tex]$ y(3) = 12 - 2(3) + 6 $[/tex]
[tex]\[ = 12 - 6 + 6 \\ = 12 \][/tex]
So, the position of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (18, 12) \)[/tex] meters.
### Step 2: Determine the Velocity
Velocity is the first derivative of the position with respect to time.
#### For x(t):
The velocity in the x-direction, [tex]\( v_x \)[/tex], is the derivative of [tex]\( x(t) \)[/tex]:
[tex]$ \frac{dx}{dt} = \frac{d}{dt}(13 + 3t - 4) \\ = 3 $[/tex]
#### For y(t):
The velocity in the y-direction, [tex]\( v_y \)[/tex], is the derivative of [tex]\( y(t) \)[/tex]:
[tex]$ \frac{dy}{dt} = \frac{d}{dt}(12 - 2t + 6) \\ = -2 $[/tex]
Therefore, the velocity of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (3, -2) \)[/tex] meters per second.
### Step 3: Determine the Acceleration
Acceleration is the second derivative of the position with respect to time or the first derivative of velocity with respect to time.
#### For x(t):
The acceleration in the x-direction, [tex]\( a_x \)[/tex], is the derivative of [tex]\( v_x \)[/tex]:
[tex]$ \frac{d^2x}{dt^2} = \frac{d}{dt}(3) \\ = 0 $[/tex]
#### For y(t):
The acceleration in the y-direction, [tex]\( a_y \)[/tex], is the derivative of [tex]\( v_y \)[/tex]:
[tex]$ \frac{d^2y}{dt^2} = \frac{d}{dt}(-2) \\ = 0 $[/tex]
Therefore, the acceleration of the particle at [tex]\( t = 3 \)[/tex] seconds is [tex]\( (0, 0) \)[/tex] meters per second squared.
### Summary
- Position at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (18, 12) \)[/tex] meters
- Velocity at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (3, -2) \)[/tex] meters per second
- Acceleration at [tex]\( t = 3 \)[/tex] seconds: [tex]\( (0, 0) \)[/tex] meters per second squared
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
