IF `t=v^2/2`, then `(-(df)/(dt))` is equal to ,
(where f is acceleration)

Options

A. `f^2`
B. `f^3`
C. `-f^3`
D. `-f^2`

Answer

To solve the problem, we start with the given relationship between \( t \) and \( v \):

1. **Given Relation**:
\[ t = \frac{v^2}{2} \]

2. **Differentiate with respect to \( t \)**:
We differentiate both sides with respect to \( t \):
\[ \frac{dt}{dt} = \frac{1}{2} \cdot 2v \cdot \frac{dv}{dt} \]
Simplifying this gives:
\[ 1 = v \cdot \frac{dv}{dt} \]

3. **Identify \( \frac{dv}{dt} \)**:
From the equation \( 1 = v \cdot \frac{dv}{dt} \), we can express \( \frac{dv}{dt} \) (which is the acceleration \( f \)):
\[ \frac{dv}{dt} = \frac{1}{v} \]
Thus, we have:
\[ f = \frac{1}{v} \]

4. **Differentiate \( vf = 1 \)**:
Now, we differentiate the equation \( vf = 1 \) with respect to \( t \):
\[ \frac{d}{dt}(vf) = 0 \]
Applying the product rule:
\[ v \frac{df}{dt} + f \frac{dv}{dt} = 0 \]

5. **Substituting \( \frac{dv}{dt} \)**:
We know \( \frac{dv}{dt} = f \), so substituting this in gives:
\[ v \frac{df}{dt} + f^2 = 0 \]

6. **Solve for \( \frac{df}{dt} \)**:
Rearranging the equation:
\[ v \frac{df}{dt} = -f^2 \]
Thus:
\[ \frac{df}{dt} = -\frac{f^2}{v} \]

7. **Substituting \( v \) in terms of \( f \)**:
From \( f = \frac{1}{v} \), we can express \( v \) as:
\[ v = \frac{1}{f} \]
Substituting this into the equation for \( \frac{df}{dt} \):
\[ \frac{df}{dt} = -\frac{f^2}{\frac{1}{f}} = -f^3 \]

8. **Finding \( -\frac{df}{dt} \)**:
Finally, we need to find \( -\frac{df}{dt} \):
\[ -\frac{df}{dt} = f^3 \]

Thus, the final answer is:
\[ -\frac{df}{dt} = f^3 \]