If `S=4 t^(3)-3t^(2)+2`, then acceleration is `42 "units" // sec ^(2)` at the time t=

Options

A. 1 sec
B. 2 sec
C. 4 sec
D. 8 sec

Answer

To find the time \( t \) when the acceleration is \( 42 \, \text{units/sec}^2 \), we will follow these steps:

### Step 1: Define the position function
Given the position function:
\[ S = 4t^3 - 3t^2 + 2 \]

### Step 2: Find the velocity function
The velocity \( v \) is the first derivative of the position function \( S \) with respect to time \( t \):
\[ v = \frac{dS}{dt} = \frac{d}{dt}(4t^3 - 3t^2 + 2) \]
Using the power rule of differentiation:
\[ v = 12t^2 - 6t \]

### Step 3: Find the acceleration function
The acceleration \( a \) is the derivative of the velocity function \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 6t) \]
Again, applying the power rule:
\[ a = 24t - 6 \]

### Step 4: Set the acceleration equal to 42
We know from the problem that the acceleration is \( 42 \, \text{units/sec}^2 \):
\[ 24t - 6 = 42 \]

### Step 5: Solve for \( t \)
Add \( 6 \) to both sides:
\[ 24t = 42 + 6 \]
\[ 24t = 48 \]
Now, divide both sides by \( 24 \):
\[ t = \frac{48}{24} = 2 \]

### Conclusion
The time \( t \) when the acceleration is \( 42 \, \text{units/sec}^2 \) is:
\[ t = 2 \, \text{seconds} \]

---