A body whose mass is 3kg performs rectilinear motion according to the formula `s = 1 + t + t^2`, where s is measured the kinetic energy `1/2 mv^2` and t in second.
Determine the kinetic energy `1/2 mv^2` of the body in `5 sec` after its start.

Options

A. `1.815 xx 10^(5)` erg
B. `1.815 xx 10^(7)` erg
C. `1.715 xx 10^(15)` erg
D. `1.815 xx 10^(8)` erg

Answer

To determine the kinetic energy of the body after 5 seconds, we will follow these steps:

### Step 1: Understand the displacement formula
The displacement \( s \) is given by the formula:
\[ s = 1 + t + t^2 \]
where \( s \) is in meters and \( t \) is in seconds.

### Step 2: Differentiate to find velocity
Velocity \( v \) is the rate of change of displacement with respect to time, which is given by:
\[ v = \frac{ds}{dt} \]
Differentiating \( s \) with respect to \( t \):
\[ \frac{ds}{dt} = 0 + 1 + 2t = 1 + 2t \]

### Step 3: Calculate the velocity at \( t = 5 \) seconds
Now, we substitute \( t = 5 \) seconds into the velocity equation:
\[ v = 1 + 2(5) = 1 + 10 = 11 \, \text{m/s} \]

### Step 4: Calculate the kinetic energy
The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the body and \( v \) is its velocity. Given that the mass \( m = 3 \, \text{kg} \) and \( v = 11 \, \text{m/s} \):
\[ KE = \frac{1}{2} \times 3 \times (11)^2 \]

### Step 5: Solve for kinetic energy
Calculating \( (11)^2 \):
\[ (11)^2 = 121 \]
Now substituting back into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 3 \times 121 = \frac{3 \times 121}{2} = \frac{363}{2} = 181.5 \, \text{J} \]

### Step 6: Convert to appropriate units (if necessary)
If required, we can convert joules to erg (1 Joule = \( 10^7 \) erg):
\[ KE = 181.5 \times 10^7 \, \text{erg} = 1.815 \times 10^5 \, \text{erg} \]

### Final Answer
Thus, the kinetic energy of the body after 5 seconds is:
\[ KE = 1.815 \times 10^5 \, \text{erg} \]

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