$A$ particle moves a distance $x$ in time $t$ according to the equation $x = (t + 5)^{-1}$. The acceleration of the particle is proportional to:

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

$A$ particle experiences a constant acceleration for $20\, s$ after starting from rest. If it travels a distance $s_1$ in the first $10\, s$ and a distance $s_2$ in the next $10\, s$,then:

The velocity $v$ of a particle as a function of its position $x$ is expressed as $v = \sqrt{c_1 - c_2 x}$,where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is

$A$ bullet moving with a velocity of $200 \, cm/s$ penetrates a wooden block and comes to rest after traversing $4 \, cm$ inside it. What velocity is needed for travelling a distance of $9 \, cm$ in the same block? (in $cm/s$)

Two cars $A$ and $B$ initially at rest are moving in the same direction with accelerations $a_1$ and $a_2$ respectively. After a certain time $t$, they achieve velocities $v_1$ and $v_2$ respectively and are separated by a distance of $50 \ m$. If $(a_1 - a_2) = 4 \ m \ s^{-2}$, then the quantity $(v_1 - v_2)$ will be: (in $m \ s^{-1}$)