A particle moving in the xy--plane has velocity ${\overrightarrow{v}}_0=v_{0x}\widehat{ı}+v_{0y}\widehat{J}$CALC at$t=0$ . It undergoes acceleration , where $b$and $c$are constants. Find an expression for the particle's velocity at a later time $t$.

The velocity of the particle at $t=0\text{s}$is ${\overrightarrow{v}}_o=v_{ox}\widehat{i}+v_{ox}\widehat{j}$.

The acceleration of the particle is $\overrightarrow{a}=bt\widehat{i}-c{v}_y\widehat{j}$.

The velocity vector of the particle is

$\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}$-

Here, $\overrightarrow{v}$is the velocity of the particle at a later time $t$, $v_x$is the horizontal component of the velocity of the particle, and $v_y$is the vertical component of the velocity of the particle.

The acceleration of the particle is,

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$-

Here, $\overrightarrow{a}$is the acceleration of the particle, and$t$ is the time.

Substitute $v_{ox}\widehat{i}+v_{oy}\widehat{j}$for$\overrightarrow{v}$and$bt\widehat{i}-c{v}_y\widehat{j}$for $\overrightarrow{a}$in the above equation to find vector equation.

$bt\widehat{i}-c{v}_y\widehat{j}=\frac{d\left(v_{\alpha }\widehat{i}+v_{os}\widehat{j}\right)}{dt}bt\widehat{i}-c{v}_y\widehat{j}=\frac{d{v}_x}{dt}\widehat{i}+\frac{d{v}_y}{dt}\widehat{j}$

The horizontal component of the velocity of the particle.

Compare the coefficient of the term $\widehat{i}$on both sides of the equation (II).

$t=\frac{d{v}_x}{dt}btdt=d{v}_x$

Integrate the above equation within the desired limit to find $v_x$.

${\int }_0^{t}btdt={\int }_{v_{ox}}^{v_x}d{v}_x{\left[\frac{b{t}^2}{2}\right]}_0^t$

$={\left[v_x\right]}_{v_{ox}}^{v_x}\left[\frac{b{t}^2}{2}-\frac{v{\left(0\right)}^2}{2}\right]$

$=\left[v_x-v_{ox}\right]v_x=v_{ox}+\frac{b{t}^2}{2}$

Thus, the horizontal component of the velocity of the particle is $v_{ox}+\frac{b{t}^2}{2}$.

Compare the coefficient of $\widehat{j}$on both sides of the equation (II).

width="175">−cvy=dvydt−cdt=dvyvy

Integrate the above equation within the desired limit

$\begin{array}{l}{\int }_0^t-cdt={\int }_{v_{oy}}^{v_y}\frac{d{v}_y}{v_y}\\ {\left[-ct\right]}_0^t={\left[\ln v_y\right]}_{v_{oy}}^{v_y}\\ \left[-ct+c\left(0\right)\right]=\left[\ln v_y-\ln v_{oy}\right]\\ v_y=v_{oy}{e}^{-ct}\end{array}$

Thus, the vertical component of the velocity of the particle is $v_{oy}{e}^{-ct}$.

Substitute $v_{ox}+\frac{b{t}^2}{2}$for $v_x$and $v_{oy}{e}^{-ct}$for $v_y$in equation (I) to find $\overrightarrow{v}$.

$\overrightarrow{v}=\left(v_{ox}+\frac{b{t}^2}{2}\right)\widehat{i}+\left(v_{oy}{e}^{-ct}\right)\widehat{j}$