Two particles $A$ and $B$ are moving in $XY$ plane. Particle $A$ moves along a line with equation $y = x$ while $B$ moves along $X$ axis such that their $X$ coordinates are always equal. If $B$ moves with a uniform speed $3\ m/s$ , the speed of $A$ is 

The co-ordinates of a particle moving in $x-y$ plane are given by :  $\mathrm{x}=2+4 \mathrm{t}, \mathrm{y}=3 \mathrm{t}+8 \mathrm{t}^2 .$ The motion of the particle is :

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 the time?

The $x$ and $y$ coordinates of the particle at any time are $x = 5t – 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The acceleration of the particle at $t = 2\, s$ is……$m/sec^2$

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