A particle is moving in a straight line. The variation of position ‘x’ as a function of time ‘t’ is given as `x = (t^3 – 6t^2 + 20t + 15)`m. The velocity of the body when its acceleration becomes zero is :

Options

A. `4 m//s`
B. `8 m//s`
C. `10 m//s`
D. `6 m//s`

Answer

To find the velocity of the particle when its acceleration becomes zero, we will follow these steps:

### Step 1: Write down the position function
The position of the particle as a function of time is given by:
\[ x(t) = t^3 - 6t^2 + 20t + 15 \]

### Step 2: Differentiate the position function to find the velocity
To find the velocity \( v(t) \), we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 20t + 15) \]
Calculating the derivative:
\[ v(t) = 3t^2 - 12t + 20 \]

### Step 3: Differentiate the velocity function to find the acceleration
Next, we differentiate the velocity function to find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20) \]
Calculating the derivative:
\[ a(t) = 6t - 12 \]

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

### Step 5: Substitute \( t \) back into the velocity function
Now that we have \( t = 2 \), we substitute this value back into the velocity function to find the velocity at this time:
\[ v(2) = 3(2^2) - 12(2) + 20 \]
Calculating:
\[ v(2) = 3(4) - 24 + 20 = 12 - 24 + 20 = 8 \text{ m/s} \]

### Final Answer:
The velocity of the body when its acceleration becomes zero is \( 8 \text{ m/s} \).

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