Derive equation of motion of body moving in two dimensions
$\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.
Derive equation of motion of body moving in two dimensions
$\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.
Consider a particle moving along $XY-$ direction with constant acceleration $\vec{a}$ and at time $t=0$, the velocity is initial velocity which is $\overrightarrow{v_{0}}$ and at time $t=t$, velocity is $\vec{v}$.
Since the particle is moving with constant acceleration, the instantaneous acceleration can be written as,
$\therefore \vec{a}=\frac{\vec{v}-\vec{v}_{0}}{t-0}$
$\therefore \vec{a}=\frac{\vec{v}-\vec{v}_{0}}{t}$
$\therefore \vec{v}=\overrightarrow{v_{0}}+\overrightarrow{a t}$
Writing this equation in terms of its components,
$v_{x}=v_{0 x}+a_{x} t$
$v_{y}=v_{0 y}+a_{y} t$
Let at $t=0$ the position vector of the particle is
given by $\overrightarrow{r_{0}}$ and at $t=t$ it is $\vec{r}$.
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