We have that
The wombats total displacement is
[tex]d_t=2100m[/tex]
The average velocity during that time is
[tex]v_{avg}=5m/s[/tex]
From the question we are told that:
initial Velocity [tex]V_1=5m/s[/tex]
initial Time [tex]T_1=4min=>240sec[/tex]
Final Velocity [tex]V_2=4m/s[/tex]
Final time [tex]T_2=3min=>180sec[/tex]
a)
Generally, the equation for displacement is mathematically given by
d_t=(Initial +final) displacement
Where
initial displacement
[tex]d_1=V_1*t_1\\\\d_1=5*240[/tex]
[tex]d_1=1200m[/tex]
Final displacement
[tex]d_2=V_2*t_2\\\\d_2=5*180[/tex]
[tex]d_2=900m[/tex]
Therefore
[tex]d_t=(Initial +final) displacement[/tex]
[tex]d_t=1200+900[/tex]
[tex]d_t=2100m[/tex]
b)
Generally, the equation for Average velocity is mathematically given by
[tex]v_{avg}=\frac{d_t}{t_1+t_2}[/tex]
[tex]v_{avg}=\frac{2100m}{240+180}[/tex]
[tex]v_{avg}=5m/s[/tex]
in conclusion
The wombats total displacement is
[tex]d_t=2100m[/tex]
The average velocity during that time is
[tex]v_{avg}=5m/s[/tex]
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