Determining Displacement and Direction of Sailboard During Wind Gust

[negation]

Homework Statement

You're sailing at 6.5ms^-1 when a wind gust hits, lasting 6.3s accelerating your board at 0.48ms^-2 at 35° to your original direction.
Find the magnitude and direction of your displacement during the gust.

The Attempt at a Solution

Attached is the magnitude of the displacement. How should I go about determining the direction of the sail board?

I ought to have been clearer in my question: I have determined the angle of displacement (6.394 degrees from the x-axis but I am unable to visualize a schematic as to what is going on)
could someone give me a leg-up?

 

[voko]

You numeric results are correct. What sort of visualization do you have in mind?

You could take a piece of paper, and draw a rectangular system of axes on it. Let at the beginning of the gust the boat be at the origin. Plot the trajectory during the gust.

 

[negation]

You numeric results are correct. What sort of visualization do you have in mind?

You could take a piece of paper, and draw a rectangular system of axes on it. Let at the beginning of the gust the boat be at the origin. Plot the trajectory during the gust.

Where should the x and y component of the velocity go? what about the angle of displacement?

 

[voko]

Do not plot the velocity. Plot the position. That will give a visualization of the trajectory.

 

[negation]

Do not plot the velocity. Plot the position. That will give a visualization of the trajectory.

Does this make sense?

 

[voko]

I would normally expect axes too be oriented differently, but that is not essential. The initial and final points are indicate correctly. But the trajectory should not be a straight line.

 

[negation]

I would normally expect axes too be oriented differently, but that is not essential. The initial and final points are indicate correctly. But the trajectory should not be a straight line.

Should it be curve due to the acceleration? If so, how should it be imposed- below the hypothenus and connecting the hypothenus?

 

[voko]

You do not have to guess. Take a number of times between zero and 6.3 seconds and compute the position at those times. Then connect them on your diagram.

 

[negation]

You do not have to guess. Take a number of times between zero and 6.3 seconds and compute the position at those times. Then connect them on your diagram.

I get it. Thanks!

 

[voko]

Here is another consideration for you. Let the y-axis be parallel to the direction of acceleration due to the gust, and x be perpendicular to it. What will the trajectory look like in these coordinates?

 

[negation]

You do not have to guess. Take a number of times between zero and 6.3 seconds and compute the position at those times. Then connect them on your diagram.

Given s = vt;

s = (48.749ms^-1i , 5.4633ms^-1j) 1.26s = 61.42m i, 6.88mj
s = (48.749ms^-1i , 5.4633ms^-1j) 2(1.26s) = 122.8474 i, 13.76mj
s = (48.749ms^-1i , 5.4633ms^-1j) 3(1.26s) = 232.16m i, 26mj
s = (48.749ms^-1i , 5.4633ms^-1j) 4(1.26s) = 309m i, 34.67mj
s = (48.749ms^-1i , 5.4633ms^-1j) 5(1.26s) = 386.946m i, 43.34mj

 

[negation]

Here is another consideration for you. Let the y-axis be parallel to the direction of acceleration due to the gust, and x be perpendicular to it. What will the trajectory look like in these coordinates?

isn't it similar to the one I just did?

 

[voko]

s = vt when there is no acceleration. Which is not the case in this problem. Besides, (48.749ms^-1i , 5.4633ms^-1j) is the final displacement, not the velocity.

 

[negation]

s = vt when there is no acceleration. Which is not the case in this problem. Besides, (48.749ms^-1i , 5.4633ms^-1j) is the final displacement, not the velocity.

What a blunder to have taken displacement x time.

If the acceleration is not constant, none of the kinematics equation can be used, isn't it?

 

[voko]

Why is the acceleration not constant? As I said, you got the correct numeric result originally. What can't you repeat the same thing, but for different times?

 

[negation]

Why is the acceleration not constant? As I said, you got the correct numeric result originally. What can't you repeat the same thing, but for different times?

Ok. I read your post wrongly. I read it as "no constant acceleration" instead of "no acceleration".

The solution is then;

ax = 0.393ms^-2
vx = 6.5ms^-1

ay = 0.2753ms^-2
vy = 0ms^-1

s = vit + 0.5at^2

s = (6.5ms^-1, 0ms^-1)([t=0,1.26,6.3]s + 0.5(0.393ms^-2,0.2753ms^-2)([t=0,1.26,6.3]^2

The above would give me the position of (i,j) from [t = 0,1.26,6.3] →from t=0 to 6.3 in segments of 1.26s

 

[voko]

Yes, that seems OK.

But pay attention to #10 as well. You do not need to compute anything there. Just think.