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, we need to find the tangential, normal, and total acceleration of a particle moving in the x-y plane with a given velocity. The velocity is given as:

\[ \vec{v} = a \hat{i} - b t \hat{j} \]

### Step 1: Find the acceleration

Acceleration \(\vec{a}\) is the rate of change of velocity, which can be calculated by differentiating the velocity with respect to time \(t\):

\[ \vec{a} = \frac{d\vec{v}}{dt} \]

Differentiating the given velocity:

\[ \vec{a} = \frac{d}{dt}(a \hat{i} - b t \hat{j}) = 0 \hat{i} - b \hat{j} = -b \hat{j} \]

### Step 2: Calculate the velocity at the given time

We need to find the velocity at the instant \(t = \frac{a \sqrt{3}}{b}\):

\[ \vec{v} = a \hat{i} - b \left(\frac{a \sqrt{3}}{b}\right) \hat{j} = a \hat{i} - a \sqrt{3} \hat{j} \]

This simplifies to:

\[ \vec{v} = a \hat{i} - a \sqrt{3} \hat{j} \]

### Step 3: Determine the angle of the velocity vector

To find the angle of the velocity vector, we can calculate the components:

- \(v_x = a\)
- \(v_y = -a \sqrt{3}\)

The angle \(\theta\) can be found using:

\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{-a \sqrt{3}}{a} = -\sqrt{3} \]

This gives us:

\[ \theta = -60^\circ \quad \text{(or equivalently, } 120^\circ \text{ in standard position)} \]

### Step 4: Calculate tangential and normal acceleration

1. **Tangential Acceleration (\(a_t\))**: This is the component of acceleration in the direction of the velocity. Since the acceleration is purely in the negative y-direction, we can find the tangential acceleration using the angle:

\[ a_t = |\vec{a}| \cos(30^\circ) = b \cdot \frac{\sqrt{3}}{2} \]

Thus, the magnitude of tangential acceleration is:

\[ a_t = \frac{\sqrt{3} b}{2} \]

2. **Normal Acceleration (\(a_n\))**: This is the component of acceleration perpendicular to the velocity. It can be calculated using:

\[ a_n = |\vec{a}| \sin(30^\circ) = b \cdot \frac{1}{2} \]

Thus, the magnitude of normal acceleration is:

\[ a_n = \frac{b}{2} \]

### Step 5: Calculate total acceleration

The total acceleration is the vector sum of tangential and normal accelerations. Since they are perpendicular, we can use the Pythagorean theorem:

\[ |\vec{a}_{total}| = \sqrt{a_t^2 + a_n^2} = \sqrt{\left(\frac{\sqrt{3} b}{2}\right)^2 + \left(\frac{b}{2}\right)^2} \]

Calculating this gives:

\[ |\vec{a}_{total}| = \sqrt{\frac{3b^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{4b^2}{4}} = \sqrt{b^2} = b \]

### Final Results

- Tangential acceleration: \(\frac{\sqrt{3} b}{2}\)
- Normal acceleration: \(\frac{b}{2}\)
- Total acceleration: \(b\)