Suppose that an object is moving in \(x-y\) plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be \(\mathbf{v}_{0}\) at time \(t=0\) and \(\mathbf{v}\) at time \(t\). Then, by definition
\(\quad \mathbf{a}=\frac{\mathbf{v}-\mathbf{v}_{0}}{t-0}=\frac{\mathbf{v}-\mathbf{v}_{0}}{t}\) Or, \(\mathbf{v}=\mathbf{v}_{\mathbf{0}} {+} \mathbf{a} t\), In terms of components :

\(v_{x}=v_{o x}+a_{x} t \) \(, v_{y}=v_{o y}+a_{y} t \)
Let us now find how the position \(\mathbf{r}\) changes with time. We follow the method used in the one-dimensional case. Let \(\mathbf{r}_{\mathrm{o}}\) and \(\mathbf{r}\) be the position vectors of the particle at time 0 and \(t\) and let the velocitles at these instants be \(\mathbf{v}_{\mathbf{o}}\) and \(\mathbf{v}\). Then, over this time interval \(t\), the average velocity is \(\left(\mathbf{v}_{0}+\mathbf{v}\right) / 2\). The displacement is the average velocity multiplied by the time interval:

\(\mathbf{r}-\mathbf{r}_{0}=\left(\frac{\mathbf{v}+\mathbf{v}_{0}}{2}\right) t=\left(\frac{\left(\mathbf{v}_{0}+\mathbf{a} t\right)+\mathbf{v}_{0}}{2}\right) t\)

\(
=\mathbf{v}_{0} t+\frac{1}{2} \mathbf{a} t^{2}
\)
Or, \(\quad \mathbf{r}=\mathbf{r}_{\mathbf{0}}+\mathbf{v}_{0} t+\frac{1}{2} \mathbf{a} t^{2}\)
Above Equation can be written in component form as
\(
\begin{aligned}
&x=x_{0}+v_{o x} t+\frac{1}{2} a_{x} t^{2} \\
&y=y_{0}+v_{o y} t+\frac{1}{2} a_{y} t^{2}
\end{aligned}
\)

Note: The velocity of the object \(\mathbf{v}_{0}\) at time \({t=0}\) is referred as \({u}\) in many reference physics books.

Example: A particle starts from origin at \(t=0\) with a velocity \(5.0 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\) and moves in \(x\) – \(y\) plane under action of a force which produces a constant acceleration of \((3.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}\). (a) What is the \(y\)-coordinate of the particle at the instant its \(x\)-coordinate is \(84 \mathrm{~m}\)? (b) What is the speed of the particle at this time?

Solution: From the position vector equation for \(\mathbf{r}_{0}=0\), the position of the particle is given by

\(\mathbf{r}=\mathbf{r}_{\mathbf{0}}+\mathbf{v}_{\mathbf{0}} t+\frac{1}{2} \mathbf{a} t^{2}\)
\(
\begin{aligned}
&\mathbf{r}(t)=\mathbf{v}_{0} t+\frac{1}{2} \mathbf{a} t^{2} \\
&=5.0 \hat{\mathbf{i}} t+(1 / 2)(3.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) t^{2} \\
&=\left(5.0 t+1.5 t^{2}\right) \hat{\mathbf{i}}+1.0 t^{2} \hat{\mathbf{j}}
\end{aligned}
\)
Therefore,
\(
\begin{aligned}
&x(t)=5.0 t+1.5 t^{2} \\
&y(t)=+1.0 t^{2}
\end{aligned}
\)
Given \(x(t)=84 \mathrm{~m}, t=\) ?
\(5.0 t+1.5 t^{2}=84 \Rightarrow t=6 \mathrm{~s}\)
At \(t=6 \mathrm{~s}, y=1.0(6)^{2}=36.0 \mathrm{~m}\)
Now, the velocity \(\mathbf{v}=\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}=(5.0+3.0 t) \hat{\mathbf{i}}+2.0 t \hat{\mathbf{j}}\)
At \(t=6 \mathrm{~s}, \quad \mathbf{v}=23.0 \hat{\mathbf{i}}+12.0 \hat{\mathbf{j}}\)
speed \(=|\mathbf{v}|=\sqrt{23^{2}+12^{2}} \cong 26 \mathrm{~m} \mathrm{~s}^{-1}\).

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