Given that the displacement of the body in meter is a function of time as follows
`x=2t^(4)+5`
The mass of the body is 2 kg. What is the increase in its kinetic energy one second after the start of motion?

Options

A. 8 J
B. 16 J
C. 32 J
D. 64 J

Explanation

To solve the problem, we need to find the increase in kinetic energy of a body after one second of motion given its displacement function. Here’s a step-by-step solution:

### Step 1: Identify the displacement function
The displacement \( x \) of the body is given by:
\[ x = 2t^4 + 5 \]

### Step 2: Differentiate the displacement to find velocity
To find the velocity \( v \), we differentiate the displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(2t^4 + 5) \]
Using the power rule of differentiation:
\[ v = 8t^3 \]

### Step 3: Calculate the velocity at \( t = 1 \) second
Now, we substitute \( t = 1 \) second into the velocity equation:
\[ v(1) = 8(1)^3 = 8 \, \text{m/s} \]

### Step 4: Determine the initial velocity
Since the body starts from rest, the initial velocity \( u \) at \( t = 0 \) seconds is:
\[ u = 0 \, \text{m/s} \]

### Step 5: Use the kinetic energy formula to find the increase in kinetic energy
The change in kinetic energy \( \Delta K \) can be calculated using the formula:
\[ \Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
Substituting the known values:
- Mass \( m = 2 \, \text{kg} \)
- Final velocity \( v = 8 \, \text{m/s} \)
- Initial velocity \( u = 0 \, \text{m/s} \)

We have:
\[ \Delta K = \frac{1}{2} \times 2 \times (8)^2 - \frac{1}{2} \times 2 \times (0)^2 \]

### Step 6: Simplify the equation
Calculating the terms:
\[ \Delta K = \frac{1}{2} \times 2 \times 64 - 0 \]
\[ \Delta K = 1 \times 64 = 64 \, \text{J} \]

### Final Answer
The increase in kinetic energy one second after the start of motion is:
\[ \Delta K = 64 \, \text{J} \]

---