A body starts moving with a velocity `v_0 = 10 ms^-1.` It experiences a retardation equal to `0.2v^2.` Its velocity after 2s is given by

Options

A. `+2 ms^-1`
B. `+4 ms^-1`
C. `-2 ms^-1`
D. `+ 6 ms^-1`

Explanation

To solve the problem step by step, we will analyze the motion of the body under the given conditions.

### Step 1: Understand the given parameters
The body starts with an initial velocity \( v_0 = 10 \, \text{m/s} \) and experiences a retardation (deceleration) that is proportional to the square of its velocity, given by \( a = -0.2v^2 \).

### Step 2: Set up the equation for acceleration
Since retardation is negative acceleration, we can express the acceleration as:
\[ a = \frac{dv}{dt} = -0.2v^2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ \frac{dv}{-0.2v^2} = dt \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( v \) from \( v_0 = 10 \, \text{m/s} \) to \( v \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( 2 \, \text{s} \):
\[ \int_{10}^{v} \frac{dv}{-0.2v^2} = \int_{0}^{2} dt \]

### Step 5: Solve the left side integral
The integral on the left side can be simplified:
\[ \int \frac{dv}{-0.2v^2} = -\frac{1}{0.2} \int v^{-2} dv = -5 \left[-\frac{1}{v}\right] = -5 \left(-\frac{1}{v} \right) = \frac{5}{v} \]
Thus, we have:
\[ \frac{5}{v} \bigg|_{10}^{v} = t \bigg|_{0}^{2} \]

### Step 6: Evaluate the limits
Substituting the limits into the equation gives:
\[ \frac{5}{v} - \frac{5}{10} = 2 - 0 \]
This simplifies to:
\[ \frac{5}{v} - 0.5 = 2 \]

### Step 7: Solve for \( v \)
Rearranging the equation to solve for \( v \):
\[ \frac{5}{v} = 2 + 0.5 = 2.5 \]
\[ v = \frac{5}{2.5} = 2 \, \text{m/s} \]

### Final Answer
The velocity of the body after 2 seconds is \( v = 2 \, \text{m/s} \).

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