A particle moves along a straight line. Its position at any instant is given by `x = 32t-(8t^3)/3` where x is in metres and t in seconds. Find the acceleration of the particle at the instant when particle is at rest.

Options

A. `-16 ms^-2`
B. `-32 ms^-2`
C. `32 ms^-2`
D. `16 ms^-2`

Explanation

To solve the problem step by step, we need to follow these procedures:

### Step 1: Write down the position function
The position of the particle is given by the equation:
\[ x(t) = 32t - \frac{8t^3}{3} \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(32t - \frac{8t^3}{3}\right) \]
Calculating the derivative:
\[ v(t) = 32 - 8t^2 \]

### Step 3: Determine when the particle is at rest
The particle is at rest when the velocity \( v(t) = 0 \):
\[ 0 = 32 - 8t^2 \]
Rearranging gives:
\[ 8t^2 = 32 \]
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \text{ seconds} \]

### Step 4: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(32 - 8t^2) \]
Calculating the derivative:
\[ a(t) = -16t \]

### Step 5: Calculate the acceleration at the instant when the particle is at rest
Now, we substitute \( t = 2 \) seconds into the acceleration function:
\[ a(2) = -16 \cdot 2 = -32 \text{ m/s}^2 \]

### Final Answer
The acceleration of the particle at the instant when it is at rest is:
\[ a = -32 \text{ m/s}^2 \]

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