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

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:

The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$. The speed of the particle at time $t$ is given by

$A$ particle starts from the origin at $t=0 \; s$ with a velocity of $10.0 \hat{j} \; m/s$ and moves in the $x-y$ plane with a constant acceleration of $(8.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
$(a)$ At what time is the $x$-coordinate of the particle $16 \; m$? What is the $y$-coordinate of the particle at that time?
$(b)$ What is the speed of the particle at that time?

When is it necessary to use vectors?

$A$ particle is moving in the $xy$-plane such that its position coordinates are given by $x = (4t + t^2) \text{ m}$ and $y = (2t + \frac{t^2}{2}) \text{ m}$, where $t$ is in seconds. What is the velocity of the particle?