A point moves along a circle having a radius `20cm` with a constant tangential acceleration `5cm//s^(2)`. How much time (in sec) is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration?

Explanation

To solve the problem step by step, we will analyze the relationship between tangential acceleration, normal acceleration, and the time required for them to be equal.

### Step 1: Understand the Definitions
1. **Tangential Acceleration (A)**: This is the acceleration that causes the speed of the object moving along the circular path to change. Given as \( A = 5 \, \text{cm/s}^2 \).
2. **Normal Acceleration (a_n)**: This is the acceleration directed towards the center of the circle, calculated using the formula \( a_n = \frac{V^2}{R} \), where \( V \) is the tangential speed and \( R \) is the radius of the circle.

### Step 2: Set Up the Equation
We need to find the time \( t \) when the normal acceleration equals the tangential acceleration:
\[ a_n = A \]
Substituting the formulas, we get:
\[ \frac{V^2}{R} = A \]

### Step 3: Express Velocity in Terms of Time
The tangential velocity \( V \) can be expressed in terms of tangential acceleration and time:
\[ V = A \cdot t \]
Substituting this into the equation for normal acceleration gives:
\[ \frac{(A \cdot t)^2}{R} = A \]

### Step 4: Substitute Known Values
Now, substitute the known values into the equation:
- \( A = 5 \, \text{cm/s}^2 \)
- \( R = 20 \, \text{cm} \)

The equation becomes:
\[ \frac{(5 \cdot t)^2}{20} = 5 \]

### Step 5: Simplify the Equation
Now simplify the equation:
1. Multiply both sides by 20:
\[ (5 \cdot t)^2 = 100 \]
2. Take the square root of both sides:
\[ 5 \cdot t = 10 \]
3. Solve for \( t \):
\[ t = \frac{10}{5} = 2 \, \text{seconds} \]

### Final Answer
The time needed after motion begins for the normal acceleration of the point to be equal to the tangential acceleration is **2 seconds**.

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