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: Let \(s = kt^n\). Then velocity \(v = nkt^{n-1}\), acceleration \(a = nk(n-1)t^{n-2}\). Compute \(nas/(n-1)\) and simplify to identify it as \(v^2\).

Step 1: \(s = kt^n\). \(v = ds/dt = nkt^{n-1}\). \(a = dv/dt = nk(n-1)t^{n-2}\).

Step 2: \(nas/(n-1) = n \cdot nk(n-1)t^{n-2} \cdot kt^n / (n-1) = n^2 k^2 t^{2n-2}\).

Step 3: \(v^2 = (nkt^{n-1})^2 = n^2 k^2 t^{2n-2}\). So \(\frac{nas}{n-1} = v^2\).

Step 4: Write the final answer. \(\boxed{v^2}\)