Respuesta :
Answer:
a) The average speed is 4.11 m/s
b) The average velocity is zero.
Explanation:
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a) The average speed is calculated as the traveled distance over the time taken to cover that distance:
a.s = d/t
Where:
a.s = average speed.
d = traveled distance.
t = time.
From A to B Wilma travels at a constant speed of 5.93 m/s. So, we can write the equation of average speed as follows:
5.93 m/s = d / t1
Where t1 is the time taken to travel the distance from A to B (d)
From B to A Wilma travels at 3.15 m/s, so the equation of average speed could be written as follows:
3.15 m/s = d / t2
Where t2 is the time taken to travel from B to A (d)
If we solve both equations for the time, we can get the total time of the trip by adding the expressions we obtain:
t1 = d / 5.93 m/s
t2 = d / 3.15 m/s
The total time of the trip is the sum of both times:
t1 + t2 = d / 5.93 m/s + d / 3.15 m/s
t1 + t2 = (3.15 m/s d + 5.93 m/s d) / (3.15 m/s)(5.93 m/s)
t1 + t2 = 9.08 m/s d / (3.15 m/s)(5.93 m/s)
The total traveled distance is 2 d and the total time is t1 + t2, so the average speed for the entire trip will be:
a.s = 2 · d / (t1 + t2)
a.s = 2 · d · ((3.15 m/s)(5.93 m/s) / 9.08 m/s d)
a.s = 4.11 m/s
The average speed is 4.11 m/s
b) The average velocity is calculated as the displacement over the elapsed time. The displacement is calculated as follows:
displacement = initial position - final position
In this case:
initial position = final position ( because Wilma returns to the starting point, there was no net displacement).
Then:
a.v = D/t
Where:
a.v = average velocity.
D = displacement
t = time.
a.v = 0/ t
a.v = 0
The average velocity is zero.
(a). Wilma's average speed over the entire trip is 4.11m/s.
(b). Wilma's average velocity over the entire trip is zero.
Let us consider that distance between A and B is x meter , time taken by Wilma to go from point A to point B is t₁ and time taken by Wilma to go from point B to point A is t₂.
(a). Given that, Wilma I. Ball walks at a constant speed of 5.93 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
Equations are,
[tex]x=5.93t_{1}\\\\x=3.15t_{2}\\\\t_{1}=\frac{x}{5.93},t_{2}=\frac{x}{3.15}[/tex]
Average speed, [tex]v=\frac{2x}{t_{1}+t_{2}}[/tex]
[tex]v=\frac{2x*18.68}{3.15x+5.93x}=\frac{37.36x}{9.08x} =4.11m/s[/tex]
(b). Since, Wilma first go from point A to point B then comeback to point A. Therefore, displacement is zero.
Average velocity [tex]=\frac{Displacement}{time}=\frac{0}{t} =0[/tex]
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