Can You Solve These Vector Questions About Wind?

[Delta2]

yes and what will be the adjacent?

 

[uniquegirl]

Sin A= √(c^2-b^2)
SinA= √(25^2-10^2)
Sin= 22.91

 

[malawi_glenn]

have you not worked with right-angled triangles and trigonometry before in basic math?

hyp = 25 (total velocity)
adj = vn (velocity component in the north direction)
opp = ve (velocity component in the east direction)

opp /hyp = sin25°
adj /hyp = cos25°

Or, perhaps this is easier

hyp = 25 (total velocity)
adj = ve (velocity component in the east direction)
opp = vn (velocity component in the north direction)

opp /hyp = sin65°
adj /hyp = cos65°

 

[Delta2]

Sin A= √(c^2-b^2)
SinA= √(25^2-10^2)
Sin= 22.91

Yes ok you can use pythagorean theorem, but you could do it as 25*cos(25)

 

[uniquegirl]

oh ok thank you! so 22.91 is the right answer? So the answer for Ve is the same as the answer for Vn, because they are same triangles?

 

[uniquegirl]

Assuming Ve is 22.91,for question b this is what I did, feel free to correct me

Question b- A strong gust of wind from the west causes the skier to accelerate to the east at 2.0 m s-2 for 1.8 s. Calculate the new velocity in the easterly direction

vi=22.65ms-1 t= 1.8s a=2.0ms^2 vf=?

vf=vi+at
vf= 22.65+2.0x1.8
vf= 26.25

 

[Delta2]

oh ok thank you! so 22.91 is the right answer? So the answer for Ve is the same as the answer for Vn, because they are same triangles?

No Ve is the opposite site , Vn is the adjacent side, the answer for Ve you calculated it as 10.24.

 

[Delta2]

Assuming Ve is 22.91,for question b this is what I did, feel free to correct me

Question b- A strong gust of wind from the west causes the skier to accelerate to the east at 2.0 m s-2 for 1.8 s. Calculate the new velocity in the easterly direction

vi=22.65ms-1 t= 1.8s a=2.0ms^2 vf=?

vf=vi+at
vf= 22.65+2.0x1.8
vf= 26.25

this would have been correct if you had used as vi=25sin(25)=10.24.

 

[uniquegirl]

Is this better? vf= 10.56+2.0x1.8 vf= 14.16

 

[uniquegirl]

C) Use your answer in (a) to calculate the size and direction of the new resultant velocity of the skier
so for direction I used tan-1=opp/hyp =10.56/25 = 0.4424 I think I did this wrong cause its such a low number but for the size, V^2= 25^2+10.56^2.

 

[malawi_glenn]

tan-1=opp/hyp

the tangent of an angle is not opp/hyp

 

[uniquegirl]

tan-1= opp/adj
= tan-1 10.56/22.91

 

[uniquegirl]

For the size, I used v=(square root ) a^2+b^2
v= (square root)22.91^2+10.56^2
would this be right?

 

[haruspex]

For the size, I used v=(square root ) a^2+b^2
v= (square root)22.91^2+10.56^2
would this be right?

The 22.91 is inaccurate. That is because in post #37 you used v_e=10 instead of 10.56. (Or, better, using v_n=25cos(25).)

Assuming you are trying part c, using 10.56m/s as the Easterly velocity here just gets you back to a net speed of 25m/s. You need to be using the new Easterly velocity calculated in post #44.

Please try to change your way of working to be entirely algebraic as far as possible, only plugging in numbers at the end. This has many advantages, such as fewer errors, easier detection of errors, improved precision and greater readability. In multipart questions, that even means not immediately using the numerical answer to one part in the next part, instead resuming the algebraic form. Sometimes that let's you do some cancellation.

When you do plug in numbers, always include the units.

To illustrate:
Part a:
v_e=v sin(25)=25m/s sin(25)= 10.56 m/s.
v_n=v cos(25)=25m/s cos(25)= 22.66 m/s.
Part b:
v_e'=v_e+at=v_e+(2m/s²)(1.8s)= 14.16 m/s.
Part c:
v'=√(v_e'²+v_n²)
etc.

Now, what about the last bit, the new direction?

 

[uniquegirl]

Cool, Thank you for all the help everyone, appreciate it