Statement-1: When velocity of a particle is zero then acceleration of particle must be zero at that instant
Statement-2: Acceleration is equal to`a= v ((dv)/(dx)) ` , where v is the velocity at that instant .

Options

A. (a)Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1
B. (b)Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1.
C. (c)Statement-1 is true, Statement-2 is false
D. (d)Statement-1 is false, Statement-2 is true

Explanation

To analyze the statements provided in the question, we will evaluate each statement step by step.

### Step 1: Evaluate Statement 1
**Statement 1:** "When the velocity of a particle is zero, then the acceleration of the particle must be zero at that instant."

- Consider a particle thrown upwards. At the highest point of its trajectory, the velocity (v) of the particle is indeed zero.
- However, even at this point, the particle experiences gravitational acceleration (g), which acts downward.
- Therefore, the acceleration is not zero; it is equal to the acceleration due to gravity (g).

**Conclusion for Statement 1:** This statement is **false**.

### Step 2: Evaluate Statement 2
**Statement 2:** "Acceleration is equal to \( a = v \frac{dv}{dx} \), where v is the velocity at that instant."

- We know that acceleration (a) can be defined as the change in velocity (dv) over the change in time (dt), i.e., \( a = \frac{dv}{dt} \).
- We can also express the change in time in terms of distance (dx) and velocity (v). Using the relationship \( dt = \frac{dx}{v} \), we can substitute this into our equation for acceleration.
- Thus, we can rewrite acceleration as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
- This shows that the equation \( a = v \frac{dv}{dx} \) is indeed valid.

**Conclusion for Statement 2:** This statement is **true**.

### Final Conclusion
- Statement 1 is **false**.
- Statement 2 is **true**.