Starting at $\mathit{t}\mathbf{=}\mathbf{0}$, a horizontal net force $\overrightarrow{\mathbf{F}}\mathbf{=}\left(0.280N/s\right)\mathit{t}\hat{\mathbf{i}}\mathbf{+}\left(-0.450N/s\right){\mathit{t}}^{\mathbf{2}}\hat{\mathbf{j}}$is applied to a box that has

an initial momentum $\overrightarrow{\mathbf{p}}\mathbf{=}\left(-3.00kg.m/s\right)\hat{\mathbf{i}}\mathbf{+}\left(4.00kg.m/s\right)\hat{\mathbf{j}}$

What is the momentum of the box at $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{s}$?

The momentum of a particle: The momentum $\overrightarrow{\mathit{p}}$ of a particle is a vector quantity equal to the product of the particle’s mass mand velocity $\overrightarrow{\mathbf{v}}$.

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{v}}$

Newton’s second law says that the net force on a particle is equal to the rate of change of the particle’s momentum.

$\stackrel{}{\mathbf{\sum }\mathbf{F}}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$

Given in the question,

Force,$\overrightarrow{F}=\left(0.280N/s\right)t\hat{i}+\left(-0.450N/s\right)t^2\hat{j}$

Initial momentum${\overrightarrow{p}}_1=\left(-3.00kg.m/s\right)\hat{i}+\left(4.00kg.m/s\right)\hat{j}$

$t_1=0$and$t_2=2.00s$

From Newton’s second law

$$\begin{gathered} \underset{}{\sum \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}} \\ d\overrightarrow{p}\sum \overrightarrow{F}dt \\ ∆\overrightarrow{p}={\int }_{t1}^{t2}\overrightarrow{F}dt \end{gathered}$$

Substituting the values

$$\begin{gathered} ∆\overrightarrow{p}={\int }_0^2\left(0.280\right)t\hat{i}+\left(-0.450\right)t^2\hat{j}dt \\ ∆\overrightarrow{p}=\left(0.280\right){\left[\frac{t^2}{2}\right]}_0^{2}i+\left(-0.450\right){\left[\frac{t^3}{3}\right]}_0^2\hat{j} \\ ∆\overrightarrow{p}=\left(0.280\right)\left[\frac{2^2}{2}-0\right]\hat{i}+\left(-0.450\right)\left[\frac{2^2}{2}-0\right]\hat{j} \\ ∆\overrightarrow{p}=0.56\hat{j}-1.2\hat{j}-1.2\hat{j}kg.m/s \end{gathered}$$

We know

$$\begin{gathered} ∆\overrightarrow{p}=\overrightarrow{p_2}-\overrightarrow{p_1} \\ 0.56\hat{i}-1.2\hat{j}={\overrightarrow{p}}_2-\left(-3.00\right)\hat{i}+\left(4.00\right)\hat{j} \\ {\overrightarrow{p}}_2=\left(-2.44\right)\hat{i}+\left(2.80\right)\hat{j}kg.m/s \end{gathered}$$

The momentum of the box at$t=2.00s$ is $\overrightarrow{p_2}=\left(-2.44kg.m/s\right)\hat{i}+\left(2.80kg.m/s\right)\hat{j}$