A particle is moving in a circular path whose portion vector is given by `barr = (acost)hat i +(asin t)hat j` then, the tangential component of acceleration at `t= (pi)/(2)` sec is

Options

A. 0
B. `am //sec^(2)`
C. `a//2 m//sec^(2)`
D. `a//4 m//sec^(2)`

Answer

To find the tangential component of acceleration for the given position vector of a particle moving in a circular path, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position Vector**:
The position vector of the particle is given by:
\[ \mathbf{r} = a \cos(t) \hat{i} + a \sin(t) \hat{j} \]

2. **Differentiate the Position Vector**:
To find the velocity vector, we differentiate the position vector with respect to time \( t \):
\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(a \cos(t) \hat{i} + a \sin(t) \hat{j}) = -a \sin(t) \hat{i} + a \cos(t) \hat{j} \]

3. **Differentiate the Velocity Vector**:
Now, we differentiate the velocity vector to find the acceleration vector:
\[ \frac{d^2\mathbf{r}}{dt^2} = \frac{d}{dt}(-a \sin(t) \hat{i} + a \cos(t) \hat{j}) = -a \cos(t) \hat{i} - a \sin(t) \hat{j} \]

4. **Evaluate at \( t = \frac{\pi}{2} \)**:
We need to find the tangential component of acceleration at \( t = \frac{\pi}{2} \):
\[ \frac{d^2\mathbf{r}}{dt^2} \bigg|_{t=\frac{\pi}{2}} = -a \cos\left(\frac{\pi}{2}\right) \hat{i} - a \sin\left(\frac{\pi}{2}\right) \hat{j} \]
Knowing that \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \):
\[ \frac{d^2\mathbf{r}}{dt^2} \bigg|_{t=\frac{\pi}{2}} = -a(0) \hat{i} - a(1) \hat{j} = -a \hat{j} \]

5. **Determine the Tangential Component of Acceleration**:
The tangential component of acceleration \( a_t \) is given by the magnitude of the acceleration vector:
\[ a_t = |-a \hat{j}| = a \text{ m/s}^2 \]

### Final Answer:
The tangential component of acceleration at \( t = \frac{\pi}{2} \) seconds is:
\[ a_t = a \text{ m/s}^2 \]