A body starts from rest and moves with uniform acceleration. What is the ratio of kinetic energies at the end of `1st , 2nd` and `3rd` seconds of its journey ?

Options

A. `1 : 8 : 27`
B. `1 : 2: 3`
C. `1 : 4: 9`
D. `3 : 2 : 1`

Explanation

To solve the problem of finding the ratio of kinetic energies at the end of the 1st, 2nd, and 3rd seconds of a body moving with uniform acceleration, we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the object and \( v \) is its velocity.

### Step 2: Determine the Velocity at Each Second
Since the body starts from rest and moves with uniform acceleration, we can use the equation of motion:
\[ v = u + at \]
Here, \( u = 0 \) (initial velocity), so:
\[ v = at \]

### Step 3: Calculate the Kinetic Energy at Each Second
1. **At the end of the 1st second (t = 1s)**:
\[ v_1 = a \cdot 1 = a \]
\[ KE_1 = \frac{1}{2} m (a)^2 = \frac{1}{2} ma^2 \]

2. **At the end of the 2nd second (t = 2s)**:
\[ v_2 = a \cdot 2 = 2a \]
\[ KE_2 = \frac{1}{2} m (2a)^2 = \frac{1}{2} m (4a^2) = 2ma^2 \]

3. **At the end of the 3rd second (t = 3s)**:
\[ v_3 = a \cdot 3 = 3a \]
\[ KE_3 = \frac{1}{2} m (3a)^2 = \frac{1}{2} m (9a^2) = \frac{9}{2} ma^2 \]

### Step 4: Find the Ratios of Kinetic Energies
Now we can find the ratio of the kinetic energies:
\[ KE_1 : KE_2 : KE_3 = \frac{1}{2} ma^2 : 2ma^2 : \frac{9}{2} ma^2 \]
To simplify, we can factor out \( \frac{1}{2} ma^2 \):
\[ = 1 : 4 : 9 \]

### Final Result
Thus, the ratio of kinetic energies at the end of the 1st, 2nd, and 3rd seconds is:
\[ \text{Ratio} = 1 : 4 : 9 \]