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

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 of a particle is given by $r = 3.0 t \hat{i} + 2.0 t^{2} \hat{j} + 5.0 \hat{k}$,where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. $(a)$ Find $v(t)$ and $a(t)$ of the particle. $(b)$ Find the magnitude and direction of $v(t)$ at $t = 1.0 \ s$.

The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 7t + 4t^2$ and $y = 5t$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The acceleration of the particle at $t = 5 \ s$ is ......... $m/s^2$.

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:

Two particles $A$ and $B$ are moving in the $XY$-plane. Their positions vary with time $t$ according to the relations:
$x_A(t) = 3t, \quad x_B(t) = 6$
$y_A(t) = t, \quad y_B(t) = 2 + 3t^2$
What is the distance between the two particles at $t = 1$?