A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate $$\frac{dM(t)}{dt} = bv^2(t)$$, where $$v(t)$$ is its instantaneous velocity. The instantaneous acceleration of the satellite is:

We consider the spaceship and the interplanetary dust as one isolated system. The important physical fact is that the dust is initially at rest in the chosen inertial frame, so before being swept up it has zero momentum. Because no external force acts on the whole system, the total linear momentum remains conserved at every instant.

The instantaneous linear momentum of the combined system after the dust is collected is simply the momentum of the spacecraft, because the newly captured dust instantly acquires the spacecraft’s velocity. Hence at time $$t$$ the momentum is

$$P(t)=M(t)\,v(t).$$

Momentum conservation in the absence of external force means

$$\frac{dP(t)}{dt}=0.$$

First we expand the derivative by using the product rule. The rule states: if $$P=uv$$ then $$\frac{dP}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}.$$ Taking $$u=M(t)$$ and $$v=v(t)$$, we obtain

$$\frac{d}{dt}\Bigl[M(t)\,v(t)\Bigr]=M(t)\,\frac{dv(t)}{dt}+v(t)\,\frac{dM(t)}{dt}=0.$$

We recognise $$\frac{dv(t)}{dt}$$ as the required instantaneous acceleration, so we denote it by $$a(t)$$. Re-writing,

$$M(t)\,a(t)+v(t)\,\frac{dM(t)}{dt}=0.$$

Solving for $$a(t)$$ gives

$$a(t)=-\frac{v(t)}{M(t)}\,\frac{dM(t)}{dt}.$$

The problem statement gives the mass-accretion rate explicitly:

$$\frac{dM(t)}{dt}=b\,v^{2}(t).$$

We now substitute this expression into the previous equation:

$$a(t)=-\frac{v(t)}{M(t)}\Bigl[b\,v^{2}(t)\Bigr].$$

Multiplying the factors of $$v(t)$$, we find

$$a(t)=-\frac{b\,v^{3}(t)}{M(t)}.$$

This is exactly the form listed in option B.

Hence, the correct answer is Option B.