Starting from the origin at time $t=0,$ with initial velocity $5 \hat{ j }\, ms ^{-1},$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{ i }+4 \hat{ j })\, ms ^{-2}$. At time $t$, its coordinates are $\left(20\, m , y _{0}\, m \right) .$ The values of $t$ and $y _{0},$ are respectively

A car travels $6\, km$ towards north at an angle of $45^o$ to the east and then travels distance of $4\, km$ towards north at an angle $135^o$ to east. How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east?

Derive equation of motion of body moving in two dimensions

$\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.

If position vector of a particle is $\left[ {(3t)\widehat i\, + \,(4{t^2})\widehat j} \right]$ , then obtain its velocity vector for $2\,s$.

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t – 5{t^2})$ meter and $x = 6t\, meter$, where $t$ is in second., the acceleration due to gravity is given by ……… $m/{\sec ^2}$