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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.
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