A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.

Options

A. `3ms^(-1)`
B. `-12ms^(-1)`
C. `42 ms^(-1)`
D. `-9ms^(-1)`

Answer

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

### Step 1: Write down the displacement equation
The displacement \( s \) of the particle is given by:
\[ s = t^3 - 6t^2 + 3t + 4 \text{ m} \]

### Step 2: Find the velocity
Velocity \( v \) is the rate of change of displacement with respect to time. Therefore, we differentiate the displacement equation with respect to \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t + 4) \]
Calculating the derivative:
\[ v = 3t^2 - 12t + 3 \]

### Step 3: Find the acceleration
Acceleration \( a \) is the rate of change of velocity with respect to time. Therefore, we differentiate the velocity equation with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 3) \]
Calculating the derivative:
\[ a = 6t - 12 \]

### Step 4: Set acceleration to zero and solve for \( t \)
We need to find the time \( t \) when the acceleration is zero:
\[ 6t - 12 = 0 \]
Solving for \( t \):
\[ 6t = 12 \implies t = 2 \text{ seconds} \]

### Step 5: Find the velocity at \( t = 2 \) seconds
Now we substitute \( t = 2 \) seconds back into the velocity equation to find the velocity at that time:
\[ v = 3(2^2) - 12(2) + 3 \]
Calculating:
\[ v = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \text{ m/s} \]

### Final Answer
The velocity when the acceleration is zero is:
\[ \boxed{-9 \text{ m/s}} \]