In the one-dimensional motion of a particle, the relation between position `x` and time `t` is given by `x^(2)+2x=t` (here `xgt0`). Choose the correct statement

Options

A. the retardatioin of the particle is `1/(4(x+1)^(3))`
B. the uniform velocity of the particle is `1/((x+1)^(3))`
C. Both are correct
D. Both are wrong

Answer

To solve the problem, we need to analyze the given equation relating position \( x \) and time \( t \):

\[ x^2 + 2x = t \]

### Step 1: Differentiate the equation with respect to time \( t \)

We start by differentiating both sides of the equation with respect to \( t \):

\[ \frac{d}{dt}(x^2 + 2x) = \frac{d}{dt}(t) \]

Using the chain rule on the left side:

\[ 2x \frac{dx}{dt} + 2 \frac{dx}{dt} = 1 \]

### Step 2: Factor out \( \frac{dx}{dt} \)

Now, we can factor out \( \frac{dx}{dt} \):

\[ \frac{dx}{dt}(2x + 2) = 1 \]

### Step 3: Solve for velocity \( v \)

Now, we can express the velocity \( v = \frac{dx}{dt} \):

\[ v = \frac{1}{2x + 2} \]

### Step 4: Find acceleration \( a \)

To find acceleration, we need to differentiate the velocity with respect to time. We can use the chain rule again:

\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]

First, we need to find \( \frac{dv}{dx} \):

\[ v = (2x + 2)^{-1} \]

Using the power rule and chain rule:

\[ \frac{dv}{dx} = -1 \cdot (2x + 2)^{-2} \cdot 2 = -\frac{2}{(2x + 2)^2} \]

Now substituting \( \frac{dx}{dt} = v \):

\[ a = \frac{dv}{dx} \cdot v = -\frac{2}{(2x + 2)^2} \cdot \frac{1}{(2x + 2)} = -\frac{2}{(2x + 2)^3} \]

### Step 5: Express retardation

Since retardation is the negative acceleration, we take the negative of \( a \):

\[ \text{Retardation} = \frac{2}{(2x + 2)^3} \]

### Conclusion

The correct statement regarding the motion of the particle is that the retardation is given by:

\[ \text{Retardation} = \frac{1}{4(x + 1)^3} \]

Thus, the first statement is correct.