Question

A particle is moving with speed \(v = b\sqrt{x}\) along positive x-axis. Calculate the speed of the particle at time \(t = \tau\) (assume that the particle is at origin at \(t = 0\)).

• A. \(\frac{b^2\tau}{\sqrt{2}}\)
• B. \(b^2\tau\)
• C. \(\frac{b^2\tau}{2}\)
• D. \(\frac{b^2\tau}{4}\)

Hint:

Write \(v=dx/dt=b\sqrt{x}\), separate variables and integrate from the origin to get x(t), then substitute in v.

Answer 1

The Correct Option is C

\(\frac{dx}{dt} = b\sqrt{x}\Rightarrow \int_0^x x^{-1/2}dx = b\int_0^t dt\).
\(2\sqrt{x} = bt\Rightarrow \sqrt{x} = \frac{bt}{2}\), so \(x = \frac{b^2t^2}{4}\).
\(v = b\sqrt{x} = b\cdot\frac{bt}{2} = \frac{b^2 t}{2}\).
At \(t=\tau\): \(v = \frac{b^2\tau}{2}\).
Answer: (3) \(\frac{b^2\tau}{2}\)