If a particle moving along a line follows the law `t=as^(2)+bs+c`, then the retardation of the particle is proportional to

Options

A. square of the velocity
B. cube of the displacement
C. cube of the velocity
D. square of the displacement

Answer

To solve the problem, we need to analyze the given equation \( t = as^2 + bs + c \) and find the relationship between the retardation of the particle and its velocity. Here’s a step-by-step solution:

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

We start with the equation:
\[ t = as^2 + bs + c \]

Differentiating both sides with respect to \( t \):
\[ \frac{dt}{dt} = \frac{d}{dt}(as^2 + bs + c) \]

This simplifies to:
\[ 1 = 2as \frac{ds}{dt} + b \frac{ds}{dt} \]

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

We can factor \( \frac{ds}{dt} \) from the right-hand side:
\[ 1 = \left(2as + b\right) \frac{ds}{dt} \]

Rearranging gives us:
\[ \frac{ds}{dt} = \frac{1}{2as + b} \]

### Step 3: Differentiate again to find acceleration

Now we need to find the acceleration \( \frac{d^2s}{dt^2} \). We differentiate \( \frac{ds}{dt} \) again with respect to \( t \):
\[ \frac{d^2s}{dt^2} = \frac{d}{dt}\left(\frac{1}{2as + b}\right) \]

Using the quotient rule:
\[ \frac{d^2s}{dt^2} = -\frac{(2a \frac{ds}{dt}) \cdot 1}{(2as + b)^2} \]

### Step 4: Substitute \( \frac{ds}{dt} \)

Substituting \( \frac{ds}{dt} = \frac{1}{2as + b} \) into the acceleration equation:
\[ \frac{d^2s}{dt^2} = -\frac{2a \cdot \frac{1}{2as + b}}{(2as + b)^2} \]

This simplifies to:
\[ \frac{d^2s}{dt^2} = -\frac{2a}{(2as + b)^3} \]

### Step 5: Relate acceleration to velocity

Recall that \( \frac{ds}{dt} \) is the velocity \( v \):
\[ v = \frac{1}{2as + b} \]

Thus, we can express \( 2as + b \) in terms of \( v \):
\[ 2as + b = \frac{1}{v} \]

Substituting this back into the acceleration equation:
\[ \frac{d^2s}{dt^2} = -2a v^3 \]

### Conclusion

The retardation (negative acceleration) is proportional to the cube of the velocity:
\[ \text{Retardation} \propto v^3 \]

Thus, the retardation of the particle is proportional to the cube of its velocity.