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:
(1) $-bv^3(t)$
(2) $\frac{-bv^3}{M(t)}$
(3) $-\frac{2bv^3}{M(t)}$
(4) $-\frac{bv^3}{2M(t)}$

A particle is moving in a straight line such that its velocity is increasing at $5 \mathrm{~m}\mathrm{~s}^{-1}$ per meter. The acceleration of the particle is $\_\_\_\_$ $\mathrm{m}\mathrm{~s}^{-2}$ at a point where its velocity is $20 \mathrm{~m}\mathrm{~s}^{-1}$.

The distance travelled by an object in time $t$ is given by $s = 2.5 t ^ { 2 }$. The instantaneous speed of the object at $t = 5 \mathrm {~s}$ will be :
(1) $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(3) $62.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $12.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position $x$ with time $t$ is given as $x = - 3 t ^ { 3 } + 18 t ^ { 2 } + 16t$, where $x$ is in m and $t$ is in s. The velocity of the particle when its acceleration becomes zero is $\_\_\_\_$ $\mathrm { m } \mathrm { s } ^ { - 1 }$.

$\vec{\mathrm{F}} = 4\mathrm{t}^{3}\hat{\mathrm{i}} - 3\mathrm{t}^{2}\hat{\mathrm{j}}, \mathrm{m} = 4\mathrm{~kg}$ at $\mathrm{t} = 0$ particle is at rest and at origin then find velocity and position at $\mathbf{t} = \mathbf{2}\mathbf{~sec}$.