$A$ body moves $6\,m$ north,$8\,m$ east,and $10\,m$ vertically upwards. What is its resultant displacement from the initial position?

$A$ particle starts from the origin at $t=0$ with a velocity $5.0 \hat{i} \; m/s$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
$(a)$ What is the $y$-coordinate of the particle at the instant its $x$-coordinate is $84 \; m$?
$(b)$ What is the speed of the particle at this time?

$A$ particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}$,where $a, b,$ and $c$ are constants. The acceleration of the particle is:

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 coordinates of a particle moving in the $XY$-plane vary with time as $x = 4t^2$ and $y = 2t$. The locus of the particle is a:

The magnitude of acceleration and velocity of a particle moving in a plane,whose position vector is $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$ at $t = 2 \text{ s}$,are,respectively: