Function of displacement x and time t is defined by equation `x=2t^(3)-3t^(2)-4t+1`. Acceleration will be zero at time

Options

A. 0.5 s
B. 2.0 s
C. 3.0 s
D. 4.0 s

Explanation

To find the time at which the acceleration is zero for the given displacement function \( x = 2t^3 - 3t^2 - 4t + 1 \), we will follow these steps:

### Step 1: Differentiate the displacement function to find velocity
The first step is to find the velocity by differentiating the displacement function \( x(t) \) with respect to time \( t \).

\[ v = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 3t^2 - 4t + 1) \]

### Step 2: Calculate the first derivative
Now, let's calculate the derivative:

\[ v = 6t^2 - 6t - 4 \]

### Step 3: Differentiate the velocity function to find acceleration
Next, we need to find the acceleration by differentiating the velocity function \( v(t) \) with respect to time \( t \).

\[ a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 6t - 4) \]

### Step 4: Calculate the second derivative
Now, let's calculate the derivative:

\[ a = 12t - 6 \]

### Step 5: Set the acceleration to zero
To find the time when the acceleration is zero, we set the acceleration function equal to zero:

\[ 12t - 6 = 0 \]

### Step 6: Solve for time \( t \)
Now, we can solve for \( t \):

\[ 12t = 6 \\ t = \frac{6}{12} \\ t = 0.5 \text{ seconds} \]

### Conclusion
Thus, the acceleration will be zero at \( t = 0.5 \) seconds.

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