A point is moving with uniform acceleration, in the eleventh and fifteenth seconds from the commencement it moves through 720 and 960 cm respectively. Its initial velocity, and the acceleration with which it moves are

Options

A. 60, 40
B. 70, 30
C. 90, 60
D. None

Answer

To solve the problem, we need to find the initial velocity (U) and the acceleration (A) of a point moving with uniform acceleration. We know the distances covered in the 11th and 15th seconds of its motion.

### Step 1: Understand the formula for distance covered in nth second
The distance covered in the nth second (S_n) can be expressed as:
\[ S_n = U + \frac{A}{2} (2n - 1) \]
where:
- \( S_n \) is the distance covered in the nth second,
- \( U \) is the initial velocity,
- \( A \) is the acceleration,
- \( n \) is the second in which the distance is measured.

### Step 2: Write equations for the 11th and 15th seconds
For the 11th second:
\[ S_{11} = U + \frac{A}{2} (2 \cdot 11 - 1) = 720 \]
This simplifies to:
\[ S_{11} = U + \frac{A}{2} (22 - 1) = U + \frac{21A}{2} = 720 \]
(Equation 1)

For the 15th second:
\[ S_{15} = U + \frac{A}{2} (2 \cdot 15 - 1) = 960 \]
This simplifies to:
\[ S_{15} = U + \frac{A}{2} (30 - 1) = U + \frac{29A}{2} = 960 \]
(Equation 2)

### Step 3: Set up the equations
From the two equations we have:
1. \( U + \frac{21A}{2} = 720 \) (Equation 1)
2. \( U + \frac{29A}{2} = 960 \) (Equation 2)

### Step 4: Subtract Equation 1 from Equation 2
Subtract Equation 1 from Equation 2 to eliminate \( U \):
\[ (U + \frac{29A}{2}) - (U + \frac{21A}{2}) = 960 - 720 \]
This simplifies to:
\[ \frac{29A}{2} - \frac{21A}{2} = 240 \]
\[ \frac{8A}{2} = 240 \]
\[ 4A = 240 \]
\[ A = 60 \text{ cm/s}^2 \]

### Step 5: Substitute A back into one of the equations to find U
Now, substitute \( A = 60 \) cm/s² back into Equation 1:
\[ U + \frac{21 \cdot 60}{2} = 720 \]
\[ U + 630 = 720 \]
\[ U = 720 - 630 \]
\[ U = 90 \text{ cm/s} \]

### Final Result
The initial velocity \( U \) is 90 cm/s and the acceleration \( A \) is 60 cm/s².