Solve Problem of Relative Movement: 52.5° South of West

[leprofece]

Please Solve this problem
View attachment 2566

Book answer
The velocity components are
x = 0.50 m/s y = (0.40 m/s)/sqrt (2) east and (0.40 m/s)/sqrt (2) south,
for a velocity relative to the Earth of 0.36 m/s, 52.5 south of west.

Ok I think these components are of an angle of 45 º
but whwn i Calculate from my grapf arctg(0,4)/0.5 = 0.8 I got a 38,39 º angle
So where do these components come from??
Thanks for your helping
taken from sears book in english edition

 

[zzephod]

Please Solve this problem
View attachment 2566

Book answer
The velocity components are
x = 0.50 m/s y = (0.40 m/s)/sqrt (2) east and (0.40 m/s)/sqrt (2) south,
for a velocity relative to the Earth of 0.36 m/s, 52.5 south of west.

Ok I think these components are of an angle of 45 º
but whwn i Calculate from my grapf arctg(0,4)/0.5 = 0.8 I got a 38,39 º angle
So where do these components come from??
Thanks for your helping
taken from sears book in english edition

Let the velocity of the boat with respect to the Earth be $\vec{v}_b$ and that of the river be $\vec{v}_r$. Then the velocity of the boat with respect to the river is $\vec{v}_{b_r}=\vec{v}_b-\vec{v}_r$

Which to me seems to give 0.455 m/s at 28.5 degrees west of south.

.

 

[leprofece]

Let the velocity of the boat with respect to the Earth be $\vec{v}_b$ and that of the river be $\vec{v}_r$. Then the velocity of the boat with respect to the river is $\vec{v}_{b_r}=\vec{v}_b-\vec{v}_r$

Which to me seems to give 0.455 m/s at 28.5 degrees west of south.

.

How do you get the angle??
arctg(o,5/0.4)?

 

[zzephod]

How do you get the angle??
arctg(o,5/0.4)?

What have you got for $\vec{v}_b$ and $\vec{v}_r$? So what do you have for $\vec{v}_{b_r}$?

The angle you are interested in is then related to $\arctan(\vec{v}_{b_r,2}/\vec{v}_{b_r,1})$, where the numeric subscripts in this last expression denote the components of $\vec{v}_{b_r}$.

.

 

[soroban]

Hello, leprofece!

3.39 A canoe has a velocity of 0.4 m/sec southeast
relative to the earth. The canoe is in on a river
that is flowing 0.5 m/s east relative to the earth.
Find the velocity (magnitude and direction of
the canoe relative to the river.

[textdraw]
A 0.5 C
o - - - - - o
\ * \
0.4 \ * \
\ * \
o - - - - - o
B D[/textdraw]
The canoe is moving from A to B.
. . [tex]\overrightarrow{AB} \:=\:\left\langle \tfrac{0.4}{\sqrt{2}},\:-\tfrac{0.4}{\sqrt{2}}\right\rangle \:=\:\langle 0.2\sqrt{2},\:-0.2\sqrt{2}\rangle[/tex]

The current is moving from A to C.
. . [tex]\overrightarrow{AC} \:=\:\langle 0.5,\:0\rangle[/tex]

The resultant is:
. . [tex]\overrightarrow{AD} \:=\:\overrightarrow{AB} + \overrightarrow{AC} \:=\:\langle 0.5+0.2\sqrt{2},\:-0.2\sqrt{2}\rangle [/tex]

Can you carry on from here?

 

[zzephod]

Hello, leprofece!

    A      0.5      C
      o - - - - - o
       \  *        \
    0.4 \     *     \
         \        *  \
          o - - - - - o
        B               D

The canoe is moving from A to B.
. . [tex]\overrightarrow{AB} \:=\:\left\langle \tfrac{0.4}{\sqrt{2}},\:-\tfrac{0.4}{\sqrt{2}}\right\rangle \:=\:\langle 0.2\sqrt{2},\:-0.2\sqrt{2}\rangle[/tex]

The current is moving from A to C.
. . [tex]\overrightarrow{AC} \:=\:\langle 0.5,\:0\rangle[/tex]

The resultant is:
. . [tex]\overrightarrow{AD} \:=\:\overrightarrow{AB} + \overrightarrow{AC} \:=\:\langle 0.5+0.2\sqrt{2},\:-0.2\sqrt{2}\rangle [/tex]

Can you carry on from here?

If the boat is moving wrt the Earth at the same velocity as the river the velocity wrt the river is 0. With your argument wrt the river its velocity is doubled.

PS repeating my arithmetic my numeric answer now agrees with the book.

.

 

[leprofece]

Hello, leprofece!


[textdraw]
A 0.5 C
o - - - - - o
\ * \
0.4 \ * \
\ * \
o - - - - - o
B D[/textdraw]
The canoe is moving from A to B.
. . [tex]\overrightarrow{AB} \:=\:\left\langle \tfrac{0.4}{\sqrt{2}},\:-\tfrac{0.4}{\sqrt{2}}\right\rangle \:=\:\langle 0.2\sqrt{2},\:-0.2\sqrt{2}\rangle[/tex]

The current is moving from A to C.
. . [tex]\overrightarrow{AC} \:=\:\langle 0.5,\:0\rangle[/tex]

The resultant is:
. . [tex]\overrightarrow{AD} \:=\:\overrightarrow{AB} + \overrightarrow{AC} \:=\:\langle 0.5+0.2\sqrt{2},\:-0.2\sqrt{2}\rangle [/tex]

Can you carry on from here?

But where do the angles come from?' I figure out the angle is 45
but is it from the figure about?? (trapeziodal)??

 

[gcfarley]

I do not understand where the \sqrt{2} is coming from

 

[MarkFL]

I do not understand where the \sqrt{2} is coming from

Consider a right isosceles triangle. If the legs have length $x$, then what length is the hypotenuse?