A particle starts from rest moves on a circle with its speed increasing at a constant rate of . Find the angle through which it `0.8 ms^(–2)` would have turned by the time its acceleration becomes `1 ms^(2)`.

Answer

To solve the problem step by step, we need to analyze the motion of the particle moving in a circular path with increasing speed.

### Step 1: Understand the given information
- The particle starts from rest, which means its initial speed \( v_0 = 0 \).
- The tangential acceleration \( a_t = 0.8 \, \text{m/s}^2 \).
- The total acceleration \( a = 1 \, \text{m/s}^2 \) when we need to find the angle \( \theta \).

### Step 2: Identify the components of acceleration
The total acceleration \( a \) of the particle can be expressed as the vector sum of tangential acceleration \( a_t \) and radial (centripetal) acceleration \( a_r \):
\[ a^2 = a_t^2 + a_r^2 \]
Given \( a = 1 \, \text{m/s}^2 \) and \( a_t = 0.8 \, \text{m/s}^2 \), we can find \( a_r \):
\[ 1^2 = (0.8)^2 + a_r^2 \]
\[ 1 = 0.64 + a_r^2 \]
\[ a_r^2 = 1 - 0.64 = 0.36 \]
\[ a_r = \sqrt{0.36} = 0.6 \, \text{m/s}^2 \]

### Step 3: Relate radial acceleration to angular velocity
The radial acceleration \( a_r \) is related to the angular velocity \( \omega \) by the formula:
\[ a_r = r \omega^2 \]
Since we do not have the radius \( r \), we will keep it as a variable for now.

### Step 4: Find the angular acceleration
The tangential acceleration \( a_t \) is related to the angular acceleration \( \alpha \) by:
\[ a_t = r \alpha \]
From this, we can express \( \alpha \):
\[ \alpha = \frac{a_t}{r} = \frac{0.8}{r} \]

### Step 5: Use the kinematic equation for angular motion
We need to find the angular displacement \( \theta \) when the angular velocity \( \omega \) is reached. We can use the kinematic equation:
\[ \omega^2 = \omega_0^2 + 2\alpha \theta \]
Since the particle starts from rest, \( \omega_0 = 0 \):
\[ \omega^2 = 2\alpha \theta \]
Substituting \( \alpha \):
\[ \omega^2 = 2 \left( \frac{0.8}{r} \right) \theta \]

### Step 6: Relate \( \omega \) to \( a_r \)
From the radial acceleration equation:
\[ 0.6 = r \omega^2 \implies \omega^2 = \frac{0.6}{r} \]
Substituting \( \omega^2 \) into the angular motion equation:
\[ \frac{0.6}{r} = 2 \left( \frac{0.8}{r} \right) \theta \]
Cancelling \( r \) (assuming \( r \neq 0 \)):
\[ 0.6 = 1.6 \theta \]
\[ \theta = \frac{0.6}{1.6} = \frac{3}{8} \, \text{radians} \]

### Final Answer
The angle through which the particle would have turned by the time its acceleration becomes \( 1 \, \text{m/s}^2 \) is:
\[ \theta = \frac{3}{8} \, \text{radians} \]