The position $x$ (in metre) of a particle moving along a straight line is given by $x=t^{3}-12 t+3$, where $t$ is time (in second). The acceleration of the particle when its velocity becomes $15 \mathrm{~ms}^{-1}$ is

A body is thrown vertically upwards with a velocity of $35 \mathrm{~ms}^{-1}$ from the ground. The ratio of the speeds of the body at times 3 s and 4 s of its motion is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A body moving with uniform acceleration, travels a distance of 25 m in the fourth second and 37 m in the sixth second. The distance covered by body in the next two seconds is

A stone is thrown vertically up from the top end of a window of height 1.8 m with a velocity of $8 \mathrm{~ms}^{-1}$. The time taken by the stone to cross the window during its downward journey is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A body is falling freely from the top of a tower of height 125 m . The distance covered by the body during the last second of its motion is $x \%$ of the height of the tower. Then, $x$ is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

The relation between time $t$ and distance $x$ of a particle is $t=a x^2+b x$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is