Question

A particle is moving on a straight line so that its distance \(s\) from a fixed point at any time \(t\) is proportional to \(t^n\). If \(v\) is the velocity and \(a\) is the acceleration of the particle at any time \(t\), then \[ \frac{nas}{n-1} = \]

• A. \(3v\)
• B. \(v^2\)
• C. \(v^3\)
• D. \(4v\) \bigskip

Hint:

Whenever \[ s\propto t^n, \] set \[ s=kt^n, \] differentiate to obtain \(v\) and \(a\), and then substitute directly. Most such questions reduce to simple power-rule differentiation.

Answer 1

The Correct Option is B

Concept: Since the distance is proportional to \(t^n\), \[ s=kt^n, \] where \(k\) is a constant. Velocity and acceleration are obtained by differentiation.

Step 1: Find the velocity. \[ s=kt^n. \] Differentiating w.r.t. \(t\), \[ v=\frac{ds}{dt} = nkt^{\,n-1}. \] \[ \cdots (1) \]

Step 2: Find the acceleration. Differentiating (1), \[ a=\frac{dv}{dt} = nk(n-1)t^{\,n-2}. \] \[ \cdots (2) \]

Step 3: Compute \(nas\). Using \[ s=kt^n \] and (2), \[ nas = n\Big[nk(n-1)t^{\,n-2}\Big] \Big[kt^n\Big]. \] \[ = n^2k^2(n-1)t^{\,2n-2}. \] Therefore, \[ \frac{nas}{n-1} = n^2k^2t^{\,2n-2}. \] \[ \cdots (3) \]

Step 4: Find \(v^2\). From (1), \[ v^2 = \left(nkt^{\,n-1}\right)^2. \] \[ = n^2k^2t^{\,2n-2}. \] \[ \cdots (4) \] From (3) and (4), \[ \frac{nas}{n-1} = v^2. \]

Step 5: Write the final answer. \[ \boxed{v^2} \]