A particle is moving with speed v = b√x along the positive x-axis. Calculate the speed of the particle at time t = τ (assume that the particle is at origin at t = 0)

The correct option is B (b²τ²)
v = b√x
dv/dt = b/(2√x) * dx/dt
Since v = dx/dt, we have
dv/dt = b/(2√x) * v = b/(2√x) * b√x = b²/2
Integrating both sides with respect to t, we get
∫dv = ∫(b²/2)dt
v = (b²/2)t + C
Since the particle is at the origin at t = 0, v = b√x = 0, so C = 0.
Therefore, v = (b²/2)t
At t = τ, v = (b²/2)τ
However, this solution is not among the given options. Let's reconsider the approach.
We have v = dx/dt = b√x
This is a separable differential equation:
dx/√x = bdt
Integrating both sides:
∫x⁻¹/² dx = ∫b dt
2√x = bt + C
Since x = 0 at t = 0, C = 0.
So, 2√x = bt
√x = bt/2
x = (b²t²)/4
Now, let's find the velocity as a function of time:
v = dx/dt = d/dt[(b²t²)/4] = (2b²t)/4 = (b²t)/2
At t = τ, v = (b²τ)/2
This is still not among the given options. Let's check the options given.
If v = b√x and dx/dt = v, then dx/dt = b√x
dx/√x = bdt
2√x = bt + C
Assuming at t = 0, x = 0, C = 0
2√x = bt
√x = bt/2
x = b²t²/4
v = dx/dt = b²t/2
At t = τ, v = b²τ/2 (Still not in the options)
Let's re-examine the question. There might be a typo in the question or the options.