A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate dMtdt=bυ2t , where υt is its instantaneous velocity. The instantaneous acceleration of the satellite is:

Thrust force = F=vdMdthere thrust force is decreasing the speed⇒Mdvdt=−vbv2⇒dvdt=−bv3Mt

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate dMtdt=bυ2t , where υt is its instantaneous velocity. The instantaneous acceleration of the satellite is:

Thrust force = F=vdMdthere thrust force is decreasing the speed⇒Mdvdt=−vbv2⇒dvdt=−bv3Mt

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

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$- \frac{b υ^{3}}{M (t)}$

$- \frac{2 b υ^{3}}{M (t)}$

$- \frac{b υ^{3}}{2 M (t)}$

$- b υ^{3} (t)$

answer is D.

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Detailed Solution

$\begin{array}{l} T h r u s t f o r c e = F = v \frac{d M}{d t} \\ h e r e t h r u s t f o r c e i s d e c r e a \sin g t h e s p e e d \\ \Rightarrow M \frac{d v}{d t} = (- v) (b v^{2}) \\ \Rightarrow \frac{d v}{d t} = - \frac{b v^{3}}{M (t)} \end{array}$