Respuesta :
Answer:
a. [tex]v_{avg}=0\ m.s^{-1}[/tex]
b. [tex]v_h=10.079\ m.s^{-1}[/tex]
c. [tex]v_c=16.282\ m.s^{-1}[/tex]
d. [tex]u_{avg}=12.451\ m.s^{-1}[/tex]
Explanation:
Given:
- time taken by the student to reach home from the class, [tex]t_h=21\ min=1260\ s[/tex]
- distance between home and physics class, [tex]s=12700\ m[/tex]
- time taken by the student to reach class from the home, [tex]t_{c}=13\ min=780\ s[/tex]
a)
We know that average velocity is given as the total displacement per unit time.
When the student goes from class to home and then back to class the the total displacement becomes zero, displacement being a vector quantity. ([tex]\Delta s=0\ m[/tex])
So, [tex]v_{avg}=0\ m.s^{-1}[/tex]
b)
Average speed of the trip from lecture hall to home:
[tex]v_h=\frac{s}{t_h}[/tex]
[tex]v_h=\frac{12700}{1260}[/tex]
[tex]v_h=10.079\ m.s^{-1}[/tex]
c)
The average speed of the girl for the trip from her home back to the lecture hall:
[tex]v_{c}=\frac{s}{t_c}[/tex]
[tex]v_c=\frac{12700}{780}[/tex]
[tex]v_c=16.282\ m.s^{-1}[/tex]
d)
Average speed is given as the total distance per unit time.
[tex]u_{avg}=\frac{2s}{t_h+t_c}[/tex]
[tex]u_{avg}=\frac{2\times 12700}{780+1260}[/tex]
[tex]u_{avg}=12.451\ m.s^{-1}[/tex]
