Respuesta :
Answer:
a) [tex] v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s[/tex]
b) [tex] \theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62 [/tex] degrees and on this case to the South of the East.
c)[tex] t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s[/tex]
d) [tex] Y = 2 m/s * 125 s = 250m[/tex]
So it would be 250 to the South
Explanation:
Part a
For this case the figure attached shows the illustration for the problem.
We know that [tex] v_y = 2 m/s[/tex] represent the velocity of the river to the south.
We have the velocity of the motorboard relative to the water and on this case is [tex]V_x= 4.8 m/s[/tex]
And we want to find the velocity of the motord board relative to the Earth [tex] v_m[/tex]
And we can find this velocity from the Pythagorean Theorem.
[tex] v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s[/tex]
Part b
We can find the direction with the following formula:
[tex] \theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62 [/tex] degrees and on this case to the South of the East.
Part c
For this case we can use the following definition
[tex] D = Vt[/tex]
The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:
[tex] t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s[/tex]
Part d
For this case we can use the same definition but now using the y compnent we have:
[tex] Y = v_y t[/tex]
And replacing we got:
[tex] Y = 2 m/s * 125 s = 250m[/tex]
So it would be 250 to the South
