Answer with explanation:
For any object having the velocity vector as
[tex]\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}+v_z\widehat{k}[/tex]
the magnitude of velocity is given by
[tex]|v|=\sqrt{v_x^2+v_y^2+v_z^2}[/tex]
For car 1 the velocity vector is
[tex]\overrightarrow{v}_1=20\widehat{i}+25\widehat{j}[/tex]
Therefore
[tex]|v_1|=\sqrt{20^2+25^2}\\\\\therefore v_1=32.0156units[/tex]
Similarly for car 2 we have
[tex]\overrightarrow{v}_2=30\widehat{i}[/tex]
Therefore
[tex]|v_2|=\sqrt{30^2}\\\\\therefore v_2=30.0units[/tex]
Comparing both the values we find that car 1 has the greater speed.