A proton in a cyclotron changes its velocity from 30km/sec north to 40km/sec east in  20sec. What is the magnitude of average acceleration (in km/s²) during this time? 

Given  $V_1=30km/s$
            $V_2=40km/s$
$\therefore$ Change in velocity  $=\left|\overline{V_2}-\overline{V_1}\right|$
                                    $=\sqrt{V_2^2+V_1^2-2V_1{V}_2\cos \theta }$
( $\therefore$When direction changes from north to east, $\theta ={90}^{\mathrm{o}}$ )
                                                  $\begin{array}{l}=\sqrt{V_2^2+V_1^2+0}\\ =\sqrt{{40}^2+{30}^2}\\ =\sqrt{1600+900}\\ =\sqrt{2500}\\ =50\mathrm{km}/\mathrm{s}\end{array}$

$\therefore$ Average acceleration  $a=\frac{\left|\overline{V_2}-\overline{V_1}\right|}{t}$

                                         $=\frac{50}{20}=2.5km/s^2$