A particle moves in the x-y plane with the velocity `bar(v)=ahati-bthatj`. At the instant `t=asqrt3//b` the magnitude of tangential, normal and total acceleration are _&_.

Answer

To solve the problem step by step, we will analyze the motion of the particle given its velocity and find the tangential, normal, and total acceleration.

### Step 1: Identify the velocity vector
The velocity of the particle is given as:
\[ \bar{v} = a \hat{i} - b t \hat{j} \]
This means that the x-component of the velocity is \(a\) and the y-component is \(-bt\).

### Step 2: Find the velocity at the specific time
We need to find the velocity at the instant \(t = \frac{a \sqrt{3}}{b}\):
\[ \bar{v} = a \hat{i} - b \left(\frac{a \sqrt{3}}{b}\right) \hat{j} = a \hat{i} - a \sqrt{3} \hat{j} \]
Thus, the velocity at this time is:
\[ \bar{v} = a \hat{i} - a \sqrt{3} \hat{j} \]

### Step 3: Calculate the acceleration vector
The acceleration \(\bar{a}\) is the time derivative of the velocity:
\[ \bar{a} = \frac{d\bar{v}}{dt} = \frac{d}{dt}(a \hat{i} - b t \hat{j}) = 0 \hat{i} - b \hat{j} = -b \hat{j} \]
So, the acceleration vector is:
\[ \bar{a} = -b \hat{j} \]

### Step 4: Determine the direction of the velocity vector
At \(t = \frac{a \sqrt{3}}{b}\), the velocity vector is:
\[ \bar{v} = a \hat{i} - a \sqrt{3} \hat{j} \]
To find the angle \(\theta\) of the velocity vector with respect to the x-axis, we can use:
\[ \tan(\theta) = \frac{\text{y-component}}{\text{x-component}} = \frac{-a \sqrt{3}}{a} = -\sqrt{3} \]
This implies \(\theta = -60^\circ\) (or \(300^\circ\) in standard position).

### Step 5: Calculate the tangential acceleration
Tangential acceleration (\(a_t\)) is the component of acceleration in the direction of the velocity. Since the acceleration vector is \(-b \hat{j}\) and the velocity vector is at an angle of \(-60^\circ\), we can find the component of acceleration in the direction of velocity:
\[ a_t = |\bar{a}| \cos(30^\circ) = b \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}b}{2} \]

### Step 6: Calculate the normal acceleration
Normal acceleration (\(a_n\)) is the component of acceleration perpendicular to the velocity. We can find this using:
\[ a_n = |\bar{a}| \sin(30^\circ) = b \cdot \frac{1}{2} = \frac{b}{2} \]

### Step 7: Calculate the total acceleration
The total acceleration is simply the magnitude of the acceleration vector:
\[ |\bar{a}| = |-b \hat{j}| = b \]

### Final Answer
Thus, the magnitudes of the tangential, normal, and total accelerations are:
- Tangential acceleration: \(\frac{\sqrt{3}b}{2}\)
- Normal acceleration: \(\frac{b}{2}\)
- Total acceleration: \(b\)