A car of mass $1000 \mathrm{kg}$ negotiates a banked curve of radius $90 \mathrm{m}$ on a frictionless road. If the banking angle is $45°$, the speed of the car is:

In the given figure, $\alpha =15 \mathrm{m} {\mathrm{s}}^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R=2.5 \mathrm{m}$ at a given instant of time. The speed of the particle is

A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) Just completes a vertical circle of diameter $AB=D$. The height $h$ is equal to

A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of '$P$' is such that it sweeps out a length $s=t^3+5$, where $s$ is in metres and $t$ is in seconds. The radius of the path is $20 \mathrm{m}$. The acceleration of 'P' when $t=2 \mathrm{s}$ is nearly-