The displacement of a particle is given by
`x=a_(0)+(a_(1)t)/(3)-(a_(2)t^(2))/(2)`
where `a_(0),a_(1)` and `a_(2)` are constants. What is its acceleration ?

Options

A. `a_(1)-a_(2)`
B. `-a_(2)`
C. `+a_(2)`
D. `a_(2)-a_(1)`

Answer

To find the acceleration of the particle given its displacement equation, we will follow these steps:

1. **Write down the displacement equation**:
\[ x = a_0 + \frac{a_1 t}{3} - \frac{a_2 t^2}{2} \]

2. **Differentiate the displacement with respect to time (t) to find the velocity (v)**:
- The velocity \( v \) is given by:
\[ v = \frac{dx}{dt} \]
- Differentiate the displacement equation:
\[ v = \frac{d}{dt} \left( a_0 + \frac{a_1 t}{3} - \frac{a_2 t^2}{2} \right) \]
- The derivative of \( a_0 \) is 0 (since it's a constant).
- The derivative of \( \frac{a_1 t}{3} \) is \( \frac{a_1}{3} \).
- The derivative of \( -\frac{a_2 t^2}{2} \) is \( -a_2 t \).
- Thus, we have:
\[ v = \frac{a_1}{3} - a_2 t \]

3. **Differentiate the velocity with respect to time (t) to find the acceleration (a)**:
- The acceleration \( a \) is given by:
\[ a = \frac{dv}{dt} \]
- Differentiate the velocity equation:
\[ a = \frac{d}{dt} \left( \frac{a_1}{3} - a_2 t \right) \]
- The derivative of \( \frac{a_1}{3} \) is 0 (since it's a constant).
- The derivative of \( -a_2 t \) is \( -a_2 \).
- Thus, we have:
\[ a = -a_2 \]

4. **Conclusion**:
- The acceleration of the particle is:
\[ a = -a_2 \]