The magnitude and direction of the velocity will be 8.60 m/sec and 54.4°.
The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
Velocity at any angle is resolved to get its component of x and y-direction.
Given data;
[tex]\rm v_x= 5 \ m/sec \\\\ v_y=7 \ m/sec[/tex]
The magnitude of the resultant velocity is;
[tex]\rm v_{net}=\sqrt{v_x^2+v_y^2}\\\\ \rm v_{net}=\sqrt{5 ^2+7^2 } \\\\ v_{net}=\sqrt{74} \\\\ v_{net}= 8.60 \ m/sec[/tex]
The direction of the velocity is;
[tex]\rm tan \theta = \frac{v_y}{v_x} \\\\ \theta = tan^{-1}(\frac{7}{5}) \\\\ \theta =54.4^0[/tex]
Hence the magnitude and direction of the velocity will be 8.60 m/sec and 54.4°.
To learn more about the velocity refer to the link ;
https://brainly.com/question/862972
#SPJ1