A body of mass 2kg at rest is supplied constant power 1J/sec, the distance travelled by body after 6 sec is

Options

A. `2sqrt6` m
B. `4sqrt m`
C. 2 sqrt3 m`
D. 6 sqrt3 m`

Answer

To solve the problem of finding the distance traveled by a body of mass 2 kg at rest when supplied with constant power of 1 J/sec over a time period of 6 seconds, we can follow these steps:

### Step 1: Understand the relationship between power, work, and kinetic energy
Power (P) is defined as the rate of doing work. The work done (W) can be expressed in terms of the change in kinetic energy (ΔKE):
\[ P = \frac{W}{t} \]
\[ W = P \cdot t \]
Since the body starts from rest, the initial kinetic energy (KE_initial) is 0. Therefore, the work done is equal to the final kinetic energy (KE_final):
\[ W = KE_{final} - KE_{initial} = KE_{final} \]

### Step 2: Calculate the work done in 6 seconds
Given that the power is 1 J/sec and the time is 6 seconds:
\[ W = P \cdot t = 1 \, \text{J/sec} \cdot 6 \, \text{sec} = 6 \, \text{J} \]

### Step 3: Relate work done to kinetic energy
Using the work-energy theorem:
\[ W = \frac{1}{2} m v^2 \]
Where:
- \( m = 2 \, \text{kg} \) (mass of the body)
- \( W = 6 \, \text{J} \)

Substituting the values:
\[ 6 = \frac{1}{2} \cdot 2 \cdot v^2 \]
\[ 6 = 1 \cdot v^2 \]
\[ v^2 = 6 \]
\[ v = \sqrt{6} \, \text{m/s} \]

### Step 4: Find the relationship between velocity and distance
The velocity of the body is not constant since it is accelerating. The power supplied is constant, which means the velocity increases over time. We can express the velocity in terms of time:
\[ v = \sqrt{t} \]

### Step 5: Set up the integral to find distance
The distance traveled (s) can be expressed as:
\[ ds = v \, dt = \sqrt{t} \, dt \]

Integrating from \( t = 0 \) to \( t = 6 \):
\[ s = \int_0^6 \sqrt{t} \, dt \]

### Step 6: Perform the integration
The integral of \( \sqrt{t} \) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]

Now, applying the limits:
\[ s = \left[ \frac{2}{3} t^{3/2} \right]_0^6 = \frac{2}{3} \left( 6^{3/2} - 0^{3/2} \right) \]
\[ s = \frac{2}{3} \cdot 6^{3/2} \]

### Step 7: Calculate \( 6^{3/2} \)
Calculating \( 6^{3/2} \):
\[ 6^{3/2} = 6 \cdot \sqrt{6} \]

Thus,
\[ s = \frac{2}{3} \cdot 6 \cdot \sqrt{6} = 4 \sqrt{6} \]

### Final Answer
The distance traveled by the body after 6 seconds is:
\[ s = 4 \sqrt{6} \, \text{meters} \]

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