The position vector of an object at any time $t$ is given by $\vec{r} = 3t^2 \hat{i} + 6t \hat{j} + \hat{k}$. Its velocity along the $y$-axis has the magnitude:

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

At time $t = 0$,a particle starts travelling from a height of $7 \, \text{cm}$ along the $z$-axis in a plane,keeping the $z$-coordinate constant. At any instant of time,its positions along the $x$ and $y$ directions are defined as $x = 3t$ and $y = 5t^3$ respectively. What will be the acceleration of the particle at $t = 1 \, \text{s}$?

The position vector of a particle changes with time according to the relation $\vec{r}(t) = 15t^2 \hat{i} + (4 - 20t^2) \hat{j}$. What is the magnitude of the acceleration at $t = 1 \ s$?

The coordinates of a moving particle at any time $t$ are given by $x = at^2$ and $y = bt^2$. The speed of the particle at any moment is

$A$ body starts from rest from the origin with an acceleration of $6 \; m/s^2$ along the $x$-axis and $8 \; m/s^2$ along the $y$-axis. Its distance from the origin after $4 \; s$ will be (in $; m$)