A car is moving on circular path of radius `100m` such that its speed is increasing at the rate of `5m//s^(2)`. At `t=0` it starts from rest. What is the radial acceleration of car at the instant it makes one complete round trip ?

Options

A. `20pi ms^(-2)`
B. `10pi ms^(-2)`
C. `5 ms^(-2)`
D. None of these

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics formulas.

### Step 1: Identify the given information
- Radius of the circular path, \( r = 100 \, \text{m} \)
- Tangential acceleration, \( a_t = 5 \, \text{m/s}^2 \)
- Initial speed, \( v_0 = 0 \, \text{m/s} \) (starts from rest)

### Step 2: Calculate the angular acceleration
The tangential acceleration \( a_t \) is related to the angular acceleration \( \alpha \) by the formula:
\[ a_t = \alpha r \]
Rearranging gives:
\[ \alpha = \frac{a_t}{r} = \frac{5 \, \text{m/s}^2}{100 \, \text{m}} = 0.05 \, \text{rad/s}^2 \]

### Step 3: Determine the time taken to complete one round
The angular displacement \( \theta \) for one complete round is \( 2\pi \) radians. Using the equation of motion for angular displacement:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Since the initial angular velocity \( \omega_0 = 0 \), we have:
\[ 2\pi = 0 + \frac{1}{2} \alpha t^2 \]
Substituting \( \alpha \):
\[ 2\pi = \frac{1}{2} (0.05) t^2 \]
Solving for \( t^2 \):
\[ t^2 = \frac{2 \cdot 2\pi}{0.05} = \frac{4\pi}{0.05} = 80\pi \]
Thus,
\[ t = \sqrt{80\pi} \]

### Step 4: Calculate the final angular velocity after one complete round
Using the angular motion equation:
\[ \omega = \omega_0 + \alpha t \]
Substituting the values:
\[ \omega = 0 + (0.05) \cdot \sqrt{80\pi} \]
\[ \omega = 0.05 \sqrt{80\pi} \]

### Step 5: Calculate the radial acceleration
The radial acceleration \( a_r \) is given by:
\[ a_r = \omega^2 r \]
Substituting \( \omega \):
\[ a_r = (0.05 \sqrt{80\pi})^2 \cdot 100 \]
Calculating \( (0.05 \sqrt{80\pi})^2 \):
\[ = 0.0025 \cdot 80\pi = 0.2\pi \]
Thus,
\[ a_r = 0.2\pi \cdot 100 = 20\pi \, \text{m/s}^2 \]
Approximating \( \pi \approx 3.14 \):
\[ a_r \approx 20 \cdot 3.14 = 62.8 \, \text{m/s}^2 \]

### Final Answer
The radial acceleration of the car at the instant it makes one complete round trip is approximately \( 62.8 \, \text{m/s}^2 \).

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