In two successive seconds a particle moves through 615 and 705 cm, respectively, assuming that it was moving with uniform acceleration, find its velocity at the commencement of the first of these two seconds and its acceleration. Find also how far it moved from rest before the commencement of the first second.

Answer

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Given:
- Distance covered in the first second, \( s_1 = 615 \) cm
- Distance covered in the second second, \( s_2 = 705 \) cm

### Step 1: Write the equations for the distances
The distance covered in the nth second can be expressed as:
\[ s_n = u + \frac{a}{2} (2n - 1) \]
where:
- \( u \) = initial velocity at the start of the first second
- \( a \) = acceleration
- \( n \) = number of seconds

For the first second (\( n = 1 \)):
\[ s_1 = u + \frac{a}{2} (2 \cdot 1 - 1) = u + \frac{a}{2} \]
Thus, we have:
\[ 615 = u + \frac{a}{2} \quad \text{(1)} \]

For the second second (\( n = 2 \)):
\[ s_2 = u + \frac{a}{2} (2 \cdot 2 - 1) = u + \frac{3a}{2} \]
Thus, we have:
\[ 705 = u + \frac{3a}{2} \quad \text{(2)} \]

### Step 2: Solve the equations
Now we have two equations:
1. \( 615 = u + \frac{a}{2} \)
2. \( 705 = u + \frac{3a}{2} \)

We can subtract equation (1) from equation (2):
\[ 705 - 615 = \left(u + \frac{3a}{2}\right) - \left(u + \frac{a}{2}\right) \]
This simplifies to:
\[ 90 = \frac{3a}{2} - \frac{a}{2} \]
\[ 90 = \frac{2a}{2} \]
\[ 90 = a \]

### Step 3: Substitute to find \( u \)
Now that we have \( a = 90 \) cm/s², we can substitute it back into equation (1) to find \( u \):
\[ 615 = u + \frac{90}{2} \]
\[ 615 = u + 45 \]
\[ u = 615 - 45 \]
\[ u = 570 \text{ cm/s} \]

### Step 4: Find the distance moved from rest before the commencement of the first second
To find the distance moved from rest before the commencement of the first second, we can use the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the particle starts from rest, we take \( u = 0 \) and \( t = 1 \) second:
\[ s = 0 \cdot 1 + \frac{1}{2} \cdot 90 \cdot (1)^2 \]
\[ s = \frac{1}{2} \cdot 90 \cdot 1 = 45 \text{ cm} \]

### Summary of Results
- Initial velocity \( u = 570 \) cm/s
- Acceleration \( a = 90 \) cm/s²
- Distance moved from rest before the commencement of the first second = 45 cm