Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Certainly! Let's work through the problem step by step.
Given:
The position function of the car is described by [tex]\( x(t) = 20t + 6t^2 \)[/tex], where [tex]\( x \)[/tex] is the distance covered at time [tex]\( t \)[/tex].
We need to calculate the instantaneous velocity and acceleration when [tex]\( t = 1 \)[/tex] second.
### Step 1: Calculate the Instantaneous Velocity
The instantaneous velocity is given by the first derivative of the position function [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex].
So,
[tex]\[ v(t) = \frac{dx(t)}{dt} \][/tex]
Given [tex]\( x(t) = 20t + 6t^2 \)[/tex], we'll differentiate this with respect to [tex]\( t \)[/tex].
[tex]\[ v(t) = \frac{d}{dt} (20t + 6t^2) \][/tex]
Using the power rule for differentiation, we get:
[tex]\[ v(t) = 20 + 12t \][/tex]
Now, we substitute [tex]\( t = 1 \)[/tex] second into this velocity equation:
[tex]\[ v(1) = 20 + 12 \cdot 1 \][/tex]
[tex]\[ v(1) = 20 + 12 \][/tex]
[tex]\[ v(1) = 32 \text{ m/s} \][/tex]
### Step 2: Calculate the Instantaneous Acceleration
The instantaneous acceleration is given by the first derivative of the velocity function [tex]\( v(t) \)[/tex] with respect to time [tex]\( t \)[/tex], which is the same as the second derivative of the position function [tex]\( x(t) \)[/tex].
So,
[tex]\[ a(t) = \frac{dv(t)}{dt} = \frac{d^2x(t)}{dt^2} \][/tex]
Given [tex]\( v(t) = 20 + 12t \)[/tex], we'll differentiate this with respect to [tex]\( t \)[/tex].
[tex]\[ a(t) = \frac{d}{dt} (20 + 12t) \][/tex]
Since the derivative of a constant is 0 and the derivative of [tex]\( 12t \)[/tex] is 12, we get:
[tex]\[ a(t) = 12 \][/tex]
In this case, the acceleration is constant and does not depend on [tex]\( t \)[/tex].
Thus, the acceleration at any time [tex]\( t \)[/tex], including at [tex]\( t = 1 \)[/tex] second, is:
[tex]\[ a(1) = 12 \text{ m/s}^2 \][/tex]
### Final Answer:
When [tex]\( t = 1 \)[/tex] second:
- The instantaneous velocity [tex]\( v = 32 \text{ m/s} \)[/tex]
- The instantaneous acceleration [tex]\( a = 12 \text{ m/s}^2 \)[/tex]
None of the options given match the correct answer.
Given:
The position function of the car is described by [tex]\( x(t) = 20t + 6t^2 \)[/tex], where [tex]\( x \)[/tex] is the distance covered at time [tex]\( t \)[/tex].
We need to calculate the instantaneous velocity and acceleration when [tex]\( t = 1 \)[/tex] second.
### Step 1: Calculate the Instantaneous Velocity
The instantaneous velocity is given by the first derivative of the position function [tex]\( x(t) \)[/tex] with respect to time [tex]\( t \)[/tex].
So,
[tex]\[ v(t) = \frac{dx(t)}{dt} \][/tex]
Given [tex]\( x(t) = 20t + 6t^2 \)[/tex], we'll differentiate this with respect to [tex]\( t \)[/tex].
[tex]\[ v(t) = \frac{d}{dt} (20t + 6t^2) \][/tex]
Using the power rule for differentiation, we get:
[tex]\[ v(t) = 20 + 12t \][/tex]
Now, we substitute [tex]\( t = 1 \)[/tex] second into this velocity equation:
[tex]\[ v(1) = 20 + 12 \cdot 1 \][/tex]
[tex]\[ v(1) = 20 + 12 \][/tex]
[tex]\[ v(1) = 32 \text{ m/s} \][/tex]
### Step 2: Calculate the Instantaneous Acceleration
The instantaneous acceleration is given by the first derivative of the velocity function [tex]\( v(t) \)[/tex] with respect to time [tex]\( t \)[/tex], which is the same as the second derivative of the position function [tex]\( x(t) \)[/tex].
So,
[tex]\[ a(t) = \frac{dv(t)}{dt} = \frac{d^2x(t)}{dt^2} \][/tex]
Given [tex]\( v(t) = 20 + 12t \)[/tex], we'll differentiate this with respect to [tex]\( t \)[/tex].
[tex]\[ a(t) = \frac{d}{dt} (20 + 12t) \][/tex]
Since the derivative of a constant is 0 and the derivative of [tex]\( 12t \)[/tex] is 12, we get:
[tex]\[ a(t) = 12 \][/tex]
In this case, the acceleration is constant and does not depend on [tex]\( t \)[/tex].
Thus, the acceleration at any time [tex]\( t \)[/tex], including at [tex]\( t = 1 \)[/tex] second, is:
[tex]\[ a(1) = 12 \text{ m/s}^2 \][/tex]
### Final Answer:
When [tex]\( t = 1 \)[/tex] second:
- The instantaneous velocity [tex]\( v = 32 \text{ m/s} \)[/tex]
- The instantaneous acceleration [tex]\( a = 12 \text{ m/s}^2 \)[/tex]
None of the options given match the correct answer.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.