A particle moving with a uniform acceleration along a straight line covers distances `a` and `b` in successive intervals of `p` and `q` second. The acceleration of the particle is

Options

A. ` (pq(p+q))/( 2(dp-aq))`
B. ` (2(aq-dp))/(pq(p-q)) `
C. ` (bp-aq)/(pq(p-q)) `
D. ` (2(bp-aq))/(pq(p+q))`

Explanation

To find the acceleration of a particle moving with uniform acceleration while covering distances \( a \) and \( b \) in successive intervals of \( p \) and \( q \) seconds, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Write the equations of motion for the first interval
For the first interval where the particle covers distance \( a \) in time \( p \), we use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ a = u p + \frac{1}{2} a p^2 \quad \text{(Equation 1)} \]

### Step 2: Write the equations of motion for the second interval
For the second interval where the particle covers distance \( b \) in time \( q \), we first need to find the initial velocity at the start of this interval. The final velocity \( v \) at the end of the first interval is given by:
\[ v = u + at \]
Substituting for \( t = p \):
\[ v = u + ap \quad \text{(Equation 2)} \]
Now, using this velocity as the initial velocity for the second interval, we write the equation for distance \( b \):
\[ b = vq + \frac{1}{2} a q^2 \]
Substituting \( v \) from Equation 2:
\[ b = (u + ap)q + \frac{1}{2} a q^2 \]
Expanding this gives:
\[ b = uq + apq + \frac{1}{2} a q^2 \quad \text{(Equation 3)} \]

### Step 3: Rearranging the equations
From Equation 1:
\[ a = up + \frac{1}{2} ap^2 \implies a - \frac{1}{2} ap^2 = up \implies a(1 - \frac{1}{2}p^2) = up \]
From Equation 3:
\[ b = uq + apq + \frac{1}{2} a q^2 \implies b - uq = apq + \frac{1}{2} a q^2 \]
Rearranging gives:
\[ b - uq = a(pq + \frac{1}{2} q^2) \quad \text{(Equation 4)} \]

### Step 4: Multiply and simplify
Now, multiply Equation 1 by \( q \) and Equation 3 by \( p \):
\[ aq = upq + \frac{1}{2} aqp^2 \quad \text{(Equation 5)} \]
\[ bp = uq + apq + \frac{1}{2} aq^2 \quad \text{(Equation 6)} \]

### Step 5: Subtract Equation 6 from Equation 5
Now, subtract Equation 6 from Equation 5:
\[ aq - bp = upq - uq + \frac{1}{2} aqp^2 - apq - \frac{1}{2} aq^2 \]
This simplifies to:
\[ aq - bp = (upq - uq) + \frac{1}{2} a(p^2 - q^2) \]

### Step 6: Solve for acceleration \( a \)
Rearranging gives:
\[ aq - bp = upq - uq + \frac{1}{2} a(p^2 - q^2) \]
Now, isolate \( a \):
\[ a(qp - \frac{1}{2}(p^2 - q^2)) = upq - uq \]
Thus,
\[ a = \frac{upq - uq}{qp - \frac{1}{2}(p^2 - q^2)} \]

### Final Expression for Acceleration
The acceleration \( a \) can be expressed as:
\[ a = \frac{2(bp - aq)}{pq(p + q)} \]