Answer:
[tex]s_2=-806m[/tex]
Explanation:
From the question we are told that
Velocity 1 [tex]v=1.34m/s \angle west[/tex]
Distance [tex]d=6.44km=>6440m[/tex]
Velocity 2 [tex]v_3=2.68m/s[/tex]
Velocity 3 [tex]v_3=0.447m/s \angle east[/tex]
Generally the equation for VT is mathematically given by
[tex]VT=v_1t_1 +V_2t_2[/tex]
[tex]VT=2.68(2400)-0.447t_2[/tex]
[tex]1.34t_1+1.34t_2=2.68(2400)-0.447t_2[/tex]
[tex]t_2=1804s[/tex]
Therefore
[tex]s_2=v_2t_2=(-0.447)(1804)[/tex]
[tex]s_2=-806m[/tex]