A car is travelling at 20 m/s on a circular road of radius 100 m. It is increasing its speed at the rate of `3" m/s"^(2)`. Its acceleration is

Options

A. `3" m/s"^(2)`
B. `4" m/s"^(2)`
C. `5" m/s"^(2)`
D. `7" m/s"^(2)`

Answer

To find the total acceleration of the car traveling on a circular road, we need to consider both the tangential acceleration and the centripetal acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Speed of the car (v) = 20 m/s
- Radius of the circular road (r) = 100 m
- Tangential acceleration (a_t) = 3 m/s²

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (a_c) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(20 \, \text{m/s})^2}{100 \, \text{m}} = \frac{400 \, \text{m}^2/\text{s}^2}{100 \, \text{m}} = 4 \, \text{m/s}^2 \]

### Step 3: Calculate the resultant acceleration
The total acceleration (a) is the vector sum of the tangential acceleration and the centripetal acceleration. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} \]
\[ a = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Step 4: Conclusion
The total acceleration of the car is \(5 \, \text{m/s}^2\).

---