IDNLearn.com provides a collaborative environment for finding and sharing answers. Ask any question and get a detailed, reliable answer from our community of experts.
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 greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.