Given that the displacement of the body in metre 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 step by step, we need to find the increase in kinetic energy of the body one second after the start of motion.

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

### Step 2: Find the velocity function
To find the velocity, we need to differentiate the displacement function 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 will calculate the velocity at \( t = 1 \) second:
\[ v(1) = 8(1)^3 = 8 \, \text{m/s} \]

### Step 4: Calculate the initial velocity at \( t = 0 \) seconds
Next, we will find the initial velocity at \( t = 0 \):
\[ v(0) = 8(0)^3 = 0 \, \text{m/s} \]

### Step 5: Calculate the increase in kinetic energy
The kinetic energy \( K \) is given by the formula:
\[ K = \frac{1}{2}mv^2 \]
The increase in kinetic energy \( \Delta K \) from \( t = 0 \) to \( t = 1 \) second is:
\[ \Delta K = K(1) - K(0) \]
Where:
- \( K(1) = \frac{1}{2} m v(1)^2 \)
- \( K(0) = \frac{1}{2} m v(0)^2 \)

Substituting the values:
- Mass \( m = 2 \, \text{kg} \)
- \( v(1) = 8 \, \text{m/s} \)
- \( v(0) = 0 \, \text{m/s} \)

Calculating \( K(1) \):
\[ K(1) = \frac{1}{2} \times 2 \times (8)^2 = 1 \times 64 = 64 \, \text{J} \]

Calculating \( K(0) \):
\[ K(0) = \frac{1}{2} \times 2 \times (0)^2 = 0 \, \text{J} \]

Thus, the increase in kinetic energy is:
\[ \Delta K = K(1) - K(0) = 64 \, \text{J} - 0 \, \text{J} = 64 \, \text{J} \]

### Final Answer
The increase in kinetic energy one second after the start of motion is **64 Joules**.

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