A particle moves in the X-Y plane with velocity `vec(v)=alphahat(i)+betathat(j)`. At `t = (alphasqrt(3))/(beta)` the magnitude of tangential, normal and total accelerations respectively are

Options

A. `(sqrt(3))/(2)beta, (beta)/(2)and beta`
B. `(beta)/(2), (beta)/(2)and beta`
C. `sqrt(2beta), sqrt(2beta)and beta`
D. None of these

Explanation

To solve the problem, we need to find the magnitudes of the tangential, normal, and total accelerations of a particle moving in the X-Y plane with a given velocity vector. Let's break it down step by step.

### Step 1: Identify the velocity vector
The velocity vector is given as:
\[ \vec{v} = \alpha \hat{i} + \beta t \hat{j} \]
where \(\alpha\) and \(\beta\) are constants, and \(t\) is time.

### Step 2: Calculate the speed
The speed \(v\) of the particle is the magnitude of the velocity vector:
\[ v = |\vec{v}| = \sqrt{(\alpha)^2 + (\beta t)^2} = \sqrt{\alpha^2 + \beta^2 t^2} \]

### Step 3: Find the tangential acceleration
The tangential acceleration \(a_t\) is defined as the rate of change of speed with respect to time:
\[ a_t = \frac{dv}{dt} \]
To find \(\frac{dv}{dt}\), we differentiate the speed:
\[ \frac{dv}{dt} = \frac{1}{2\sqrt{\alpha^2 + \beta^2 t^2}} \cdot (0 + \beta^2 \cdot 2t) = \frac{\beta^2 t}{\sqrt{\alpha^2 + \beta^2 t^2}} \]
Now, we need to evaluate this at \(t = \frac{\alpha \sqrt{3}}{\beta}\):
\[ a_t = \frac{\beta^2 \left(\frac{\alpha \sqrt{3}}{\beta}\right)}{\sqrt{\alpha^2 + \beta^2 \left(\frac{\alpha \sqrt{3}}{\beta}\right)^2}} = \frac{\beta^2 \frac{\alpha \sqrt{3}}{\beta}}{\sqrt{\alpha^2 + \frac{3\alpha^2}{\beta^2} \beta^2}} = \frac{\beta \alpha \sqrt{3}}{\sqrt{4\alpha^2}} = \frac{\beta \alpha \sqrt{3}}{2\alpha} = \frac{\beta \sqrt{3}}{2} \]

### Step 4: Find the normal acceleration
The normal acceleration \(a_n\) can be found using the relationship:
\[ a_n = \sqrt{a^2 - a_t^2} \]
where \(a\) is the total acceleration. The total acceleration \(a\) is the derivative of the velocity vector:
\[ \vec{a} = \frac{d\vec{v}}{dt} = 0 \hat{i} + \beta \hat{j} = \beta \hat{j} \]
Thus, the magnitude of the total acceleration is:
\[ a = |\vec{a}| = \beta \]
Now, we can find \(a_n\):
\[ a_n = \sqrt{\beta^2 - \left(\frac{\beta \sqrt{3}}{2}\right)^2} = \sqrt{\beta^2 - \frac{3\beta^2}{4}} = \sqrt{\frac{4\beta^2}{4} - \frac{3\beta^2}{4}} = \sqrt{\frac{\beta^2}{4}} = \frac{\beta}{2} \]

### Step 5: Summary of results
- Tangential acceleration \(a_t = \frac{\beta \sqrt{3}}{2}\)
- Normal acceleration \(a_n = \frac{\beta}{2}\)
- Total acceleration \(a = \beta\)

### Final Answer:
The magnitudes of the tangential, normal, and total accelerations respectively are:
\[ \frac{\beta \sqrt{3}}{2}, \frac{\beta}{2}, \beta \]