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[13] The distance covered by a car at a time [tex]\( t \)[/tex] is given by [tex]\( x = 20t + 6t^2 \)[/tex]. Calculate the instantaneous velocity and acceleration when [tex]\( t = 1 \, \text{s} \)[/tex].

A. [tex]\( v = 43 \, \text{m/s}, \, a = 71 \, \text{m/s}^2 \)[/tex]

B. [tex]\( v = 44 \, \text{m/s}, \, a = 72 \, \text{m/s}^2 \)[/tex]

C. [tex]\( v = 45 \, \text{m/s}, \, a = 73 \, \text{m/s}^2 \)[/tex]

D. [tex]\( v = 46 \, \text{m/s}, \, a = 74 \, \text{m/s}^2 \)[/tex]

Space for your working:
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Answer :

GinnyAnswer
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.

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