A spaceship in space sweeps stationary interplanetary dust.

Question:

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass

increases at a rate $\frac{\mathrm{dM}(\mathrm{t})}{\mathrm{dt}}=\mathrm{bv}^{2}(\mathrm{t})$, where $\mathrm{v}(\mathrm{t})$

is its instantaneous velocity. The instantaneous acceleration of the satellite is:

  1. $-\frac{2 b v^{3}}{M(t)}$ 

  2. $-\frac{\mathrm{bv}^{3}}{2 \mathrm{M}(\mathrm{t})}$

  3. $-b v^{3}(t)$

  4. $-\frac{\mathrm{bv}^{3}}{\mathrm{M}(\mathrm{t})}$


Correct Option: , 4

Solution:

$\frac{\mathrm{dm}(\mathrm{t})}{\mathrm{dt}}=\mathrm{bv}^{2}$'

$\mathrm{F}_{\text {thast }}=\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}$'

Force on statellile $=-\overrightarrow{\mathrm{v}} \frac{\mathrm{dm}(\mathrm{t})}{\mathrm{dt}}$

$\mathrm{M}(\mathrm{t}) \mathrm{a}=-\mathrm{v}\left(\mathrm{bv}^{2}\right)$

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

 

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