A point moves in the plane so that its tangential acceleration `w_tau=a`, and its normal acceleration `w_n=bt^4`, where a and b are positive constants, and t is time. At the moment `t=0` the point was at rest. Find how the curvature radius R of the point's trajectory and the total acceleration w depend on the distance covered s.

Answer

To solve the problem step by step, we will analyze the given information about the motion of the point and derive the required expressions for the curvature radius \( R \) and the total acceleration \( w \).

### Step 1: Understand the given information
We have:
- Tangential acceleration \( w_{\tau} = a \)
- Normal acceleration \( w_n = bt^4 \)
- Initial conditions: At \( t = 0 \), the point is at rest.

### Step 2: Find the velocity as a function of time
Since the tangential acceleration is constant, we can use the formula for velocity:
\[ v(t) = v_0 + w_{\tau} t \]
Given that \( v_0 = 0 \) (the point is at rest), we have:
\[ v(t) = at \]

### Step 3: Find the distance covered as a function of time
Using the equation of motion for distance:
\[ s(t) = v_0 t + \frac{1}{2} w_{\tau} t^2 \]
Substituting \( v_0 = 0 \):
\[ s(t) = \frac{1}{2} a t^2 \]

### Step 4: Express time \( t \) in terms of distance \( s \)
From the equation \( s(t) = \frac{1}{2} a t^2 \), we can solve for \( t \):
\[ t^2 = \frac{2s}{a} \quad \Rightarrow \quad t = \sqrt{\frac{2s}{a}} \]

### Step 5: Find the normal acceleration in terms of distance \( s \)
We know that the normal acceleration is given by:
\[ w_n = \frac{v^2}{R} \]
Substituting \( v(t) = at \):
\[ w_n = \frac{(at)^2}{R} = \frac{a^2 t^2}{R} \]
Now substituting \( t^2 = \frac{2s}{a} \):
\[ w_n = \frac{a^2 \left(\frac{2s}{a}\right)}{R} = \frac{2as}{R} \]
We also know from the problem statement that:
\[ w_n = bt^4 \]
Substituting \( t^4 = \left(\sqrt{\frac{2s}{a}}\right)^4 = \frac{4s^2}{a^2} \):
\[ w_n = b \cdot \frac{4s^2}{a^2} \]

### Step 6: Equate the two expressions for normal acceleration
Setting the two expressions for \( w_n \) equal to each other:
\[ \frac{2as}{R} = b \cdot \frac{4s^2}{a^2} \]
Rearranging gives:
\[ R = \frac{2as^2}{b \cdot 4s} = \frac{as}{2b} \]

### Step 7: Find the total acceleration \( w \)
The total acceleration \( w \) is given by:
\[ w = \sqrt{w_{\tau}^2 + w_n^2} \]
Substituting the known values:
\[ w = \sqrt{a^2 + \left(bt^4\right)^2} \]
Substituting \( t^4 = \frac{4s^2}{a^2} \):
\[ w = \sqrt{a^2 + b^2 \left(\frac{4s^2}{a^2}\right)^2} = \sqrt{a^2 + \frac{16b^2s^4}{a^4}} \]

### Final Results
1. The curvature radius \( R \) of the point's trajectory is:
\[ R = \frac{as}{2b} \]
2. The total acceleration \( w \) depends on the distance \( s \) as:
\[ w = \sqrt{a^2 + \frac{16b^2s^4}{a^4}} \]