A car runs around a curve of radius 10m at a constant speed of `10ms^(-1)`. Consider the time interval for which car covers a curve of `120^(@)` are:-

Options

A. Resultant change in velocity of car `10sqrt3 ms^(-1)`
B. instantaneous acceleration of car is `10ms^(-2)`
C. Average acceleration of car is `(5)/(24)ms^(-2)`
D. Instantaneous and average acceleration are same for the given period of motion.

Answer

To solve the problem of a car running around a curve of radius 10 m at a constant speed of 10 m/s while covering an angle of 120 degrees, we will follow these steps:

### Step 1: Calculate the distance covered by the car
The distance \( s \) covered by the car while turning through an angle \( \theta \) in radians can be calculated using the formula:
\[ s = r \theta \]
where \( r \) is the radius and \( \theta \) is the angle in radians.

First, convert the angle from degrees to radians:
\[ \theta = 120^\circ = \frac{120 \times \pi}{180} = \frac{2\pi}{3} \text{ radians} \]

Now, substitute the values into the formula:
\[ s = 10 \, \text{m} \times \frac{2\pi}{3} = \frac{20\pi}{3} \, \text{m} \]

### Step 2: Calculate the time taken to cover this distance
The time \( t \) taken to cover the distance at a constant speed \( v \) is given by:
\[ t = \frac{s}{v} \]
Substituting the values we have:
\[ t = \frac{\frac{20\pi}{3}}{10} = \frac{2\pi}{3} \, \text{s} \]

### Final Answer
The time interval for which the car covers a curve of 120 degrees is:
\[ t = \frac{2\pi}{3} \, \text{s} \approx 2.094 \, \text{s} \]

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