A particle of mass $1 \mathrm{kg}$ has a velocity of $2 \mathrm{m}/\mathrm{s}$. A constant force of $2N$ acts on the particle for $1 \mathrm{s}$ in a direction perpendicular to its initial velocity. Find the velocity and displacement of the particle at the end of $1 \mathrm{s}$.

Force acting on the particle is constant. Hence, Acceleration of the particle will also remain constant. 

$a=\frac{F}{m}=\frac{2}{1}=2 \mathrm{m}/{\mathrm{s}}^2$

Since, acceleration is constant. We can apply

$v=u+at$

and $s=ut+\frac{1}{2}a{t}^2$

Refer figure (a).

Here, $u$and $at$ are two mutually perpendicular vectors. So,

$$\begin{gathered} \left|v\right|=\sqrt{\left(\right|\mathbf{u}|{)}^2+(\left|\mathbf{a}t\right|{)}^2} \\ =\sqrt{(2{)}^2+(2{)}^2} \\ =2\sqrt{2}m/s \\ \alpha ={\tan }^{-1}\frac{\left|\mathrm{a}t\right|}{\left|\mathrm{u}\right|}={\tan }^{-1}\left(\frac{2}{2}\right) \\ ={\tan }^{-1}\left(1\right)=45° \end{gathered}$$

The Velocity of the particle at the end of 1s is $2\sqrt{2}m/s$ at an angle of ${45}^o$with its initial velocity. 

Refer figure (b). $s=ut+\frac{1}{2}a{t}^2$

Here $ut and \frac{1}{2}a{t}^2$ are also two mutually perpendicular vectors. So,

$$\begin{gathered} \left|\mathbf{s}\right|=\sqrt{\left(\right|\mathbf{u}t|{)}^2+{\left(\left|\frac{1}{2}\mathbf{a}{t}^2\right|\right)}^2} \\ =\sqrt{(2{)}^2+(1{)}^2} \\ =\sqrt{5}m \end{gathered}$$

and $$\begin{gathered} \beta ={\tan }^{-1}\frac{\left|\frac{1}{2}{\mathbf{at}}^2\right|}{\left|\mathbf{u}t\right|} \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \end{gathered}$$

Thus, displacement of the particle at the end of 1 s is $\sqrt{5}m$ at an angle of ${\tan }^{-1}\left(\frac{1}{2}\right)$ from its initial velocity.