Respuesta :
Answer:
Case:1 a) average speed [tex]v_{avg}=4.1144\ m.s^{-1}[/tex]
Since the body returns to its initial position hence the displacement is zero so velocity is also zero.
CASE: 2 a) average speed [tex]v_{avg}=3.7093\ m.s^{-1}[/tex]
Since the body returns to its initial position hence the displacement is zero so velocity is also zero.
Explanation:
Given:
Case: 1
speed of Wilma from point A to B, [tex]v_{ab}=5.93\ m.s^{-1}[/tex]
speed of Wilma from point B to A, [tex]v_{ba}=3.15\ m.s^{-1}[/tex]
Let the displacement from A to B be x meter.
Then the total displacement = 2x meter
time taken in going from A to B:
[tex]t_{ab}=\frac{x}{5.93}\ s[/tex]
time taken in going from B to A:
[tex]t_{ba}=\frac{x}{3.15} \ s[/tex]
Wilma's average speed over the entire trip:
[tex]v_{avg}=\frac{2x}{t_{ab}+t_{ba}}[/tex]
[tex]v_{avg}=2x\div (\frac{x}{5.93}+\frac{x}{3.15})[/tex]
[tex]v_{avg}=4.1144\ m.s^{-1}[/tex]
- Since the body returns to its initial position hence the displacement is zero so velocity is also zero.
CASE: 2
speed of Wilma from point A to B, [tex]v_{ab}=4.51\ m.s^{-1}[/tex]
speed of Wilma from point B to A, [tex]v_{ba}=3.15\ m.s^{-1}[/tex]
Let the displacement from A to B be x meter.
Then the total displacement = 2x meter
time taken in going from A to B:
[tex]t_{ab}=\frac{x}{4.51}\ s[/tex]
time taken in going from B to A:
[tex]t_{ba}=\frac{x}{3.15} \ s[/tex]
Wilma's average speed over the entire trip:
[tex]v_{avg}=\frac{2x}{t_{ab}+t_{ba}}[/tex]
[tex]v_{avg}=2x\div (\frac{x}{4.51}+\frac{x}{3.15})[/tex]
[tex]v_{avg}=3.7093\ m.s^{-1}[/tex]
- Since the body returns to its initial position hence the displacement is zero so velocity is also zero.
