$A$ particle moves along a straight line according to the law $s = \frac{1}{3} t^3 - 3 t^2 + 9 t + 17$,where $s$ is in metre and $t$ is in second. Its velocity decreases in

If the displacement of a particle at a point is given by $s = 3t^{2} - 12t + 14$, then the displacement of the particle when its velocity becomes zero is (in $\text{ units}$)

The $x$-coordinate changes on the curve $y=3x^5+15x-8$ at the rate of $\frac{1}{5} \text{ units/sec}$. If $A(x_1, y_1)$ and $B(x_2, y_2)$ are the points on the curve at which the $y$-coordinate changes at the rate of $6 \text{ units/sec}$, then the slope of $AB$ is:

Moving along the $x$-axis are two points with $x = 10 + 6t$ and $x = 3 + t^2$. The speed with which they are receding from each other at the time of encounter is ........... $cm/sec$. ($x$ is in $cm$ and $t$ is in seconds)

If the distance $s$ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2t$,then the acceleration at $t = 0$ is

If the distance travelled by a point in time $t$ is $s = 180t - 16t^2$,then the rate of change in velocity is ......... $unit$.