The position of a particle as a function of time is described by relation `x = 3t-3 t^2 + t ^(3),` where the quantities are expressed in SI units. If mass of the particle be 10 kg, the work done in first three seconds is

Options

A. 10 J
B. 30 J
C. 300 J
D. 675 J

Explanation

To solve the problem step by step, we will follow these instructions:

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

### Step 2: Find the velocity function
To find the velocity, we need to differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t - 3t^2 + t^3) \]
Using the power rule of differentiation:
\[ v(t) = 3 - 6t + 3t^2 \]

### Step 3: Calculate the velocity at \( t = 0 \) seconds
Substituting \( t = 0 \) into the velocity function:
\[ v(0) = 3 - 6(0) + 3(0)^2 = 3 \, \text{m/s} \]

### Step 4: Calculate the velocity at \( t = 3 \) seconds
Now, substituting \( t = 3 \) into the velocity function:
\[ v(3) = 3 - 6(3) + 3(3)^2 \]
Calculating this gives:
\[ v(3) = 3 - 18 + 27 = 12 \, \text{m/s} \]

### Step 5: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = \frac{1}{2} m (v^2 - u^2) \]
where \( m = 10 \, \text{kg} \), \( v = v(3) = 12 \, \text{m/s} \), and \( u = v(0) = 3 \, \text{m/s} \).

Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 10 \times (12^2 - 3^2) \]
Calculating \( 12^2 \) and \( 3^2 \):
\[ 12^2 = 144 \]
\[ 3^2 = 9 \]
Now substituting these values:
\[ \Delta KE = \frac{1}{2} \times 10 \times (144 - 9) \]
\[ \Delta KE = \frac{1}{2} \times 10 \times 135 \]
\[ \Delta KE = 5 \times 135 = 675 \, \text{J} \]

### Step 6: Conclusion
The work done on the particle in the first three seconds is equal to the change in kinetic energy:
\[ \text{Work Done} = 675 \, \text{J} \]

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