Let            ${\mathbf{r}}_1\left(t\right)=3t\hat{\mathbf{i}}+4t^2\hat{\mathbf{j}}$

and           ${\mathbf{r}}_2\left(t\right)=4t^2\hat{\mathbf{i}}+3\hat{\mathbf{j}}$

represent the positions of particles 1 and 2, respectively, as function of time t, r₁ (t) and r₂ (t) are in metre and t in second. The relative speed of the two particles at the instant t = 1s, will be            

 

Given, ${\mathbf{r}}_1\left(t\right)=3t\hat{\mathbf{i}}+4t^2\hat{\mathbf{j}}$

$\therefore$       $\frac{d{\mathbf{r}}_1}{dt}=3\hat{\mathbf{i}}+8\hat{\mathbf{j}}$

At         t = 1s,

             ${\mathbf{v}}_1=\frac{d{\mathbf{r}}_1}{dt}=3\hat{\mathbf{i}}+8\hat{\mathbf{j}}$

Again,   ${\mathbf{r}}_2\left(t\right)=4t^2\hat{\mathbf{i}}+3\hat{\mathbf{j}}$

$\Rightarrow \frac{d{\mathbf{r}}_2}{dt}=8t\hat{\mathbf{i}}+3\hat{\mathbf{j}}$

At                 t = 1s,

                    ${\mathbf{v}}_2=\frac{d{\mathbf{r}}_{\mathbf{2}}}{dt}=8\hat{\mathbf{i}}+3\hat{\mathbf{j}}$

Relative speed, ${\mathrm{v}}_{\mathrm{rel}} = {\mathrm{v}}_2 - {\mathrm{v}}_1$

                                $=(8\hat{\mathbf{i}}+3\hat{\mathbf{j}})-(3\hat{\mathbf{i}}+8\hat{\mathbf{j}})$

                                $=5\hat{\mathbf{i}}-5\hat{\mathbf{j}}$

$\therefore$                  $\left|{\mathbf{v}}_{\text{rel }}\right|=\sqrt{(5{)}^2+(-5{)}^2}$

                               $= 5\sqrt{2} \mathrm{m}/\mathrm{s}$