A particle moving along the x axis has position given by x = `(24t - 2.0t^3)` m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

Options

A. `24 ms^(-1) `
B. zero
C. `12 ms^(-1) `
D. `48 ms^(-1) `

Answer

To solve the problem step by step, we will find the acceleration of a particle when its velocity is zero.

### Step 1: Write down the position function
The position of the particle is given by the equation:
\[ x(t) = 24t - 2.0t^3 \]

### Step 2: Find the velocity function
Velocity is the derivative of the position function with respect to time. Therefore, we differentiate \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(24t - 2.0t^3) \]
Calculating the derivative:
\[ v(t) = 24 - 6.0t^2 \]

### Step 3: Set the velocity to zero to find the time when velocity is zero
To find the time when the velocity is zero, set \( v(t) = 0 \):
\[ 24 - 6.0t^2 = 0 \]
Rearranging gives:
\[ 6.0t^2 = 24 \]
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \text{ seconds} \]

### Step 4: Find the acceleration function
Acceleration is the derivative of the velocity function with respect to time. We differentiate \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(24 - 6.0t^2) \]
Calculating the derivative:
\[ a(t) = -12.0t \]

### Step 5: Calculate the acceleration at \( t = 2 \) seconds
Now, we substitute \( t = 2 \) seconds into the acceleration function:
\[ a(2) = -12.0(2) = -24.0 \text{ m/s}^2 \]

### Step 6: Find the magnitude of the acceleration
The magnitude of acceleration is the absolute value:
\[ |a| = 24.0 \text{ m/s}^2 \]

### Final Answer
The magnitude of the acceleration of the particle when its velocity is zero is:
\[ \boxed{24.0 \text{ m/s}^2} \]

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