
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate \[\dfrac{{\left[ {{\rm{dM}}\left( {\rm{t}} \right)} \right]}}{{{\rm{dt}}}} = {\rm{d}}{{\rm{v}}^2}({\rm{t}})\], where ${\rm{v}}\left( {\rm{t}} \right)$ is its instantaneous velocity. The instantaneous acceleration of the satellite is:
(a) $ - {\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)$
(b) $\dfrac{{ - 2{\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)}}{{{\rm{M}}\left( {\rm{t}} \right)}}$
(c)$\dfrac{{ - {\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)}}{{{\rm{M}}\left( {\rm{t}} \right)}}$
(d) $\dfrac{{ - 2{\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)}}{{{\rm{2M}}\left( {\rm{t}} \right)}}$
Answer
164.1k+ views
Hint: Force is an external agent that has the power to alter a body's resting or moving position. It has a direction and a magnitude. The direction of the force is referred to as the force's direction, and the point of application is where the force is exerted. We also need to determine the satellite's current instantaneous acceleration.
Formula used:
${\rm{F}} = {\rm{ma}}$
Complete answer:
According to the question \[\dfrac{{\left[ {{\rm{dM}}\left( {\rm{t}} \right)} \right]}}{{{\rm{dt}}}} = {\rm{d}}{{\rm{v}}^2}({\rm{t}})\] is the rate at which mass increases and instantaneous velocity is ${\rm{v}}\left( {\rm{t}} \right)$.
As we know that force is the product of mass and acceleration. We know that from the second law of motion as the object goes above the mass decreases so applying Newton’s second law of motion we can say that,
$F = \dfrac{{dp}}{{dt}} = \dfrac{d}{{dt}}\left( {Mv} \right)$
After differentiation
$F = M\dfrac{{dv}}{{dt}} + v\dfrac{{dM}}{{dt}}$
As the satellite will go above the thrust will be got canceled by force, therefore $F = 0$
$\begin{array}{c}
- M\dfrac{{dv}}{{dt}} = v\dfrac{{dM}}{{dt}}\\
- Ma = v\dfrac{{dM}}{{dt}}\\
a = \dfrac{{v\left( {b{v^2}} \right)}}{M} = \dfrac{{ - {\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)}}{{{\rm{M}}\left( {\rm{t}} \right)}}
\end{array}$
Hence option (c) is the correct option.
Note: The acceleration of an item caused by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object, according to Newton's second law of motion. The same method used to calculate instantaneous velocity can also be used to calculate instantaneous acceleration or acceleration at a single moment in time.
Formula used:
${\rm{F}} = {\rm{ma}}$
Complete answer:
According to the question \[\dfrac{{\left[ {{\rm{dM}}\left( {\rm{t}} \right)} \right]}}{{{\rm{dt}}}} = {\rm{d}}{{\rm{v}}^2}({\rm{t}})\] is the rate at which mass increases and instantaneous velocity is ${\rm{v}}\left( {\rm{t}} \right)$.
As we know that force is the product of mass and acceleration. We know that from the second law of motion as the object goes above the mass decreases so applying Newton’s second law of motion we can say that,
$F = \dfrac{{dp}}{{dt}} = \dfrac{d}{{dt}}\left( {Mv} \right)$
After differentiation
$F = M\dfrac{{dv}}{{dt}} + v\dfrac{{dM}}{{dt}}$
As the satellite will go above the thrust will be got canceled by force, therefore $F = 0$
$\begin{array}{c}
- M\dfrac{{dv}}{{dt}} = v\dfrac{{dM}}{{dt}}\\
- Ma = v\dfrac{{dM}}{{dt}}\\
a = \dfrac{{v\left( {b{v^2}} \right)}}{M} = \dfrac{{ - {\rm{b}}{{\rm{v}}^3}\left( {\rm{t}} \right)}}{{{\rm{M}}\left( {\rm{t}} \right)}}
\end{array}$
Hence option (c) is the correct option.
Note: The acceleration of an item caused by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object, according to Newton's second law of motion. The same method used to calculate instantaneous velocity can also be used to calculate instantaneous acceleration or acceleration at a single moment in time.
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