Respuesta :
Answer:
[tex]v_{avg}=77.1874\ km.hr^{-1}[/tex]
[tex]v'_{avg}=77.1874\ km.hr^{-1}[/tex]
[tex]v_a=77.1874\ km.hr^{-1}[/tex]
[tex]s=0\ km[/tex] because the person finally return to the initial position.
Explanation:
Given:
velocity during the first half, [tex]u_1=65\ km.hr^{-1}[/tex]
velocity during the second half, [tex]u_2=95\ km.hr^{-1}[/tex]
velocity during the first half of onward journey, [tex]v_1=65\ km.hr^{-1}[/tex]
velocity during the second half of onward journey, [tex]v_2=95\ km.hr^{-1}[/tex]
Let the total displacement and distance between the two points be 2x meters.
So, the time taken for the first half of the journey:
[tex]t_1=\frac{x}{65} \ hr[/tex]
The time taken for the second half of the journey:
[tex]t_2=\frac{x}{95} \ hr[/tex]
Time taken for the first half of onward journey:
[tex]t'_1=\frac{x}{65} \ hr[/tex]
Time taken for the first half of onward journey:
[tex]t'_2=\frac{x}{95}\ hr[/tex]
average speed going down to St. Louis:
[tex]v_{avg}=\frac{Total\ distance}{total\ time}[/tex]
[tex]v_{avg}=\frac{2x}{t_1+t_2}[/tex]
[tex]v_{avg}=\frac{2x}{(\frac{x}{65}+\frac{x}{95}) }[/tex]
[tex]v_{avg}=77.1874\ km.hr^{-1}[/tex]
Similarly, average speed coming back from St. Louis:
[tex]v'_{avg}=\frac{2x}{t'_1+t'_2}[/tex]
[tex]v'_{avg}=\frac{2x}{(\frac{x}{65}+\frac{x}{95}) }[/tex]
[tex]v'_{avg}=77.1874\ km.hr^{-1}[/tex]
Now the total average speed of the trip:
[tex]v_a=\frac{2x+2x}{(t_1+t_2+t'_1+t'_2)}[/tex]
[tex]v_a=\frac{4x}{(\frac{x}{65} +\frac{x}{95} +\frac{x}{65} +\frac{x}{95} )}[/tex]
[tex]v_a=77.1874\ km.hr^{-1}[/tex]
Total displacement during the entire trip:
[tex]s=0\ km[/tex] because the person finally return to the initial position.
