A body accelerates uniformly from rest to a velocity 1 `ms^(-1)` in 15 seconds.The kinetic energy of the body will be `(2)/(9) j` when `t` is equal to [Take mass of body as 1 kg]

Options

A. 4s
B. 8s
C. 10s
D. 12s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the acceleration of the body
The body accelerates uniformly from rest (initial velocity \( u = 0 \, \text{m/s} \)) to a final velocity \( v = 1 \, \text{m/s} \) in \( t = 15 \, \text{s} \).

Using the formula for acceleration:
\[ a = \frac{v - u}{t} \]
Substituting the values:
\[ a = \frac{1 \, \text{m/s} - 0 \, \text{m/s}}{15 \, \text{s}} = \frac{1}{15} \, \text{m/s}^2 \]

### Step 2: Calculate the velocity at time \( t \)
The velocity \( v(t) \) at any time \( t \) can be calculated using the formula:
\[ v(t) = u + at \]
Substituting the known values:
\[ v(t) = 0 + \left(\frac{1}{15} \, \text{m/s}^2\right) \cdot t = \frac{t}{15} \, \text{m/s} \]

### Step 3: Calculate the kinetic energy at time \( t \)
The kinetic energy \( KE \) of the body is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Given the mass \( m = 1 \, \text{kg} \):
\[ KE = \frac{1}{2} \cdot 1 \cdot \left(\frac{t}{15}\right)^2 = \frac{1}{2} \cdot \frac{t^2}{225} = \frac{t^2}{450} \, \text{J} \]

### Step 4: Set the kinetic energy equal to \( \frac{2}{9} \, \text{J} \)
We need to find the time \( t \) when the kinetic energy is \( \frac{2}{9} \, \text{J} \):
\[ \frac{t^2}{450} = \frac{2}{9} \]

### Step 5: Solve for \( t^2 \)
Cross-multiplying gives:
\[ t^2 = \frac{2}{9} \cdot 450 \]
Calculating the right side:
\[ t^2 = \frac{2 \cdot 450}{9} = \frac{900}{9} = 100 \]
Thus,
\[ t = \sqrt{100} = 10 \, \text{s} \]

### Conclusion
The time \( t \) at which the kinetic energy of the body is \( \frac{2}{9} \, \text{J} \) is \( 10 \, \text{s} \).

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