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Step-by-Step Solution
Step 1: Understand the problem
We have a spaceship traveling in space that is sweeping up interplanetary dust. As it collects dust, its mass changes with time. The rate of change of mass is given by
$ \frac{dM(t)}{dt} = b\,v^2(t) $,
where $v(t)$ is the instantaneous velocity of the spaceship and $b$ is a constant. We need to find the instantaneous acceleration $a(t)$ of the spaceship under these conditions.
Step 2: Apply the principle of conservation of momentum
In free space, there are no external forces other than the interaction due to collecting dust. When the spaceship picks up dust that was initially at rest relative to it, the effective force is associated with the momentum change from the added mass.
Step 3: Express the force on the spaceship
The momentum of the system at any time is $ M(t)\,v(t) $. The force can be computed from
$ F = \frac{d}{dt} \big( M(t)\,v(t) \big) $.
However, since dust is being swept in (and effectively accelerates from rest to the spaceship's velocity), the thrust-like force on the spaceship points opposite to its motion. One commonly used form for variable-mass systems is:
$ \vec{F} = -\,\vec{v}\,\frac{dM(t)}{dt}, $
because the incoming mass contributes negative momentum to the spaceship in the reference frame of the spaceship.
Step 4: Substitute the rate of mass change
We know $ \frac{dM(t)}{dt} = b\,v^2(t) $. Therefore, if $F$ is along the negative $v$ direction:
$ F = M(t)\,a(t) = -\,v(t)\,\Big(b\,v^2(t)\Big) = -\,b\,v^3(t). $
Step 5: Solve for the acceleration
From $ M(t)\,a(t) = -\,b\,v^3(t), $ we have
$$
a(t) = \frac{F}{M(t)} = -\,\frac{b\,v^3(t)}{M(t)}.
$$
This shows that the acceleration is inversely proportional to the instantaneous mass and depends on the cube of the spacecraft’s velocity.
Final Answer
$ a(t) = -\,\frac{b\,v^3(t)}{M(t)}. $
