A particle is moving in a straight line with constant acceleration. If x,y and z be the distances described by a particle during the pth, qth and rth second respectively, prove that (q-r)x+(r-p)y+(p-q)z=0

Explanation

To prove the equation \( (q - r)x + (r - p)y + (p - q)z = 0 \), where \( x, y, z \) are the distances covered by a particle during the \( p \)-th, \( q \)-th, and \( r \)-th seconds respectively, we will use the equations of motion under constant acceleration.

### Step 1: Write the distance equations for each second
The distance covered by the particle in the \( n \)-th second can be given by the formula:
\[ s_n = u + \frac{a}{2}(2n - 1) \]
where \( u \) is the initial velocity and \( a \) is the constant acceleration.
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