A particle moves in a circle in such a way that, its tangential deceleration is numerically equal to its radial acceleration. If the initial velocity of the particle is `V_(0)`, find the variation of its velocity with time

Options

A. `(V_(0))/(1+ (V_(0)t)/(R))`
B. `(V_(0)t)/(1+ (V_(0)t)/(R))`
C. `(V_(0))/(t+ (V_(0)t)/(R))`
D. None of above

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular path with given conditions. Here's a step-by-step solution:

### Step 1: Understand the given conditions
The problem states that the tangential deceleration (negative acceleration) is numerically equal to the radial acceleration. We denote:
- Tangential deceleration: \( a_t = -\alpha \) (where \( \alpha \) is a positive constant)
- Radial (centripetal) acceleration: \( a_r = \frac{v^2}{r} \)

Given that \( a_t = a_r \), we have:
\[ \alpha = \frac{v^2}{r} \]

### Step 2: Relate deceleration to velocity change
The tangential deceleration can also be expressed in terms of the change in velocity:
\[ a_t = \frac{dv}{dt} = -\alpha \]
Thus, we can write:
\[ \frac{dv}{dt} = -\frac{v^2}{r} \]

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

### Step 4: Integrate both sides
Now we integrate both sides. The left side will be integrated from \( v_0 \) to \( v \) (the initial and final velocities), and the right side will be integrated from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v^2} = -\int_{0}^{t} \frac{dt}{r} \]

The left side integrates to:
\[ -\frac{1}{v} \bigg|_{v_0}^{v} = -\frac{1}{v} + \frac{1}{v_0} \]
The right side integrates to:
\[ -\frac{t}{r} \bigg|_{0}^{t} = -\frac{t}{r} \]

### Step 5: Set the integrals equal to each other
Setting the results of the integrals equal gives:
\[ -\frac{1}{v} + \frac{1}{v_0} = -\frac{t}{r} \]

### Step 6: Solve for \( \frac{1}{v} \)
Rearranging this equation, we find:
\[ \frac{1}{v} = \frac{1}{v_0} + \frac{t}{r} \]

### Step 7: Invert to find \( v \)
Taking the reciprocal of both sides gives:
\[ v = \frac{1}{\frac{1}{v_0} + \frac{t}{r}} \]

### Step 8: Simplify the expression
To simplify this expression, we can combine the terms:
\[ v = \frac{v_0 r}{r + v_0 t} \]

### Final Result
Thus, the variation of the velocity \( v \) with time \( t \) is given by:
\[ v(t) = \frac{v_0 r}{r + v_0 t} \]