A body is moving in a circle with a speed of `1 ms^-1`. This speed increases at a constant rate of `2 ms^-1` every second. Assume that the radius of the circle described is `25 m`. The total accleration of the body after `2 s` is.

Options

A. `2 m s^-2`
B. `25 ms^-2`
C. `sqrt(5) ms^-2`
D. `sqrt(7) ms^-2`

Explanation

To find the total acceleration of a body moving in a circle after 2 seconds, we need to calculate both the tangential acceleration and the centripetal acceleration. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial speed (u) = 1 m/s
- Rate of increase of speed (tangential acceleration, a_t) = 2 m/s²
- Radius of the circle (r) = 25 m
- Time (t) = 2 s

### Step 2: Calculate the final speed after 2 seconds
Using the first equation of motion:
\[ V = U + a_t \cdot t \]
Where:
- V = final speed
- U = initial speed
- a_t = tangential acceleration
- t = time

Substituting the values:
\[ V = 1 \, \text{m/s} + (2 \, \text{m/s}^2 \cdot 2 \, \text{s}) \]
\[ V = 1 \, \text{m/s} + 4 \, \text{m/s} \]
\[ V = 5 \, \text{m/s} \]

### Step 3: Calculate the centripetal acceleration
Centripetal acceleration (a_c) is given by the formula:
\[ a_c = \frac{V^2}{r} \]
Substituting the values:
\[ a_c = \frac{(5 \, \text{m/s})^2}{25 \, \text{m}} \]
\[ a_c = \frac{25 \, \text{m}^2/\text{s}^2}{25 \, \text{m}} \]
\[ a_c = 1 \, \text{m/s}^2 \]

### Step 4: Calculate the total acceleration
The total acceleration (a) is the vector sum of the tangential acceleration and 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:
\[ a = \sqrt{(2 \, \text{m/s}^2)^2 + (1 \, \text{m/s}^2)^2} \]
\[ a = \sqrt{4 + 1} \]
\[ a = \sqrt{5} \, \text{m/s}^2 \]

### Final Answer
The total acceleration of the body after 2 seconds is:
\[ a = \sqrt{5} \, \text{m/s}^2 \]

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