2-d Kinematics (Need to verify solution)

  • #1

Homework Statement


A fish swimming in a horizontal plane has a velocity of 4.123 m/s [N 75.96deg E] in the ocean with a position vector of 10.770 m [S 68.20 deg E] relative to a stationary rock at the shore. After the fish swims with constant acceleration for 20.00 s, its velocity is 20.62 m/s [S 75.96 deg E].

(a) Calculate the acceleration.
(b) Where is the fish at 25.00 s and in what direction is it moving?

Homework Equations


AT = AX + AY

N, E (+)

V2x = -(sin75.96)(20.62)
V1x = (cos14.04)(4.123)
V2y = (cos75.96)(20.62)
V1y = (sin14.04)(4.123)

The Attempt at a Solution


(a)
x-dir
Ax = (-V2x - V1x) / t
= ( -(sin75.96)(20.62) - (cos14.04)(4.123) ) / 20
= -1.2002 m/s2

y-dir
Ay = (V2y - V1y) / t
= ( (cos75.96)(20.62) - (sin14.04)(4.123) ) / 20
= 0.2001 m/s 2

At = sqrt(ax2 + ay2)
= 1.217 m/s2

tan(theta) = ax/ay
theta = 81deg

Therefore aT = 1.217 m/s2 [S 81deg E]

(b) For b, I know I would split up DT into two components the same way I did above.
I would use the ax, ay, t=25, v1x, v1y, d1x and d1y that I already have

For example,
dx = v1x (t) + 1/2 at^2
dx = (cos14.04)(4.123) + 1/2(-1.2001)^2

This is asking the distance at 25s.. The problem states that the position vector is 10.770m[ S 68.26 E] meaning that I would have to add this to the equation, I think.

I would take the x-component of the position vector (sin68.20)(10.770) and add it to the dx equation I wrote above, like so:
dx = (sin68.20(10.770) + (cos14.04)(4.123) + 1/2(-1.2001)^2

then do the same for dy and then find the dt value + magnitude.

Did I do both parts correctly?
 
Last edited:

Answers and Replies

  • #2
gneill
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I think that there's a small problem with your calculation of the vector components of v2, perhaps in the interpretation of the angle notation?
 
  • #3
Well =going over my solution I'm not sure but here's the diagram I drew of the V2 vector+components

2dkzfw6.png
 
  • #4
gneill
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If the direction is [S 75.96 deg E], doesn't that make it 75.96 degrees east of south, which would make the x-component positive (eastward) and the y-component negative (southward)?
 
  • #5
If you were to draw it like that, the direction would become [E 14.04 deg S], which is equivalent to [S 75.96 deg E], no?
 
  • #6
gneill
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If you were to draw it like that, the direction would become [E 14.04 deg S], which is equivalent to [S 75.96 deg E], no?

I suppose it would.
 
  • #7
So would you say my solution is correct for Parts A and B ?
Especially wondering about whether my idea for B is correct
 
  • #8
gneill
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No. I think you have a problem with the components of your vectors, as I said. If you are equating the North direction with the +Y axis, and the East direction with the +X direction, shouldn't V2 have a positive x component and a negative y component?

You have written:

N, E (+)

V2x = -(sin75.96)(20.62) ----> = -20.00 m/s
V2y = (cos75.96)(20.62) ----> = 5.00 m/s

V1x = (cos14.04)(4.123) ----> = 4.00 m/s
V1y = (sin14.04)(4.123) ----> = 1.00 m/s

I agree with the V1 components, but have a problem with the V2 components. I think that their signs should be reversed.
 
  • #9
Ahh yes you're right! I see that now.
Do you think I have the correct idea for part b, when it comes to finding out where the fish will be at t=25s?

Also, provided that is correct... How would I find the 'direction in which it is moving' ?
Would I take the said DT value and use its components and SOH CAH TOA (forming a triangle) to find direction and magnitude?
 
  • #10
gneill
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Yes, simply take the initial position and velocity vectors and apply the "usual" equation of motion to them:

d(t) = do + vot + 1/2)at^2

where all the variables are vectors (except time of course). You can do the x and y components separately.

To find the direction it's going at time t, you'll want to find the velocity vector at that time:

v(t) = vo + at

From the components of the velocity vector you can extract your direction information.
 
  • #11
The do you have written there stands for the x or y component of the position vector that is given, right?

So for example.. finding my x component first, I would have
d(t) = (sin68.20)(10.770) + (cos14.04)(4.123)t + (1/2)at^2
 
  • #12
gneill
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The do you have written there stands for the x or y component of the position vector that is given, right?

So for example.. finding my x component first, I would have
d(t) = (sin68.20)(10.770) + (cos14.04)(4.123)t + (1/2)at^2

Actually it stands for both. Yes, do it component wise and you'll be fine. Remember that the acceleration, a, has components too!
 
  • #13
My teacher told me that I am supposed to add the dO afterwards...This didnt make sense to me. It doesnt matter whether you include it in the d(t) equation right?
 
  • #14
gneill
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My teacher told me that I am supposed to add the dO afterwards...This didnt make sense to me. It doesnt matter whether you include it in the d(t) equation right?

The order in which you add things isn't so important as knowing what has to be added! If you want to find the final position, it has to be with respect to something. The initial position (do) gives you that starting place reference (which, in turn, is apparently referenced to some rock at the shore).

So, yes, it matters that it be included when determining the final position.
 
  • #15
OK I ended up with "At 25s, the fish is at d2= 367.3m [S 78.57 deg E]
after doing x and y components separately

For the next part (direction its moving in), you said
"To find the direction it's going at time t, you'll want to find the velocity vector at that time:

v(t) = vo + at

From the components of the velocity vector you can extract your direction information."

But I dont understand why thats necessary. Isnt the direction in the answer I got for d2? The fish is going in the southeast direction?
 
  • #16
Nevermind, I understand now. Anyway, I solved for the components of the velocity vector and I got that V1 = 112.0m/s [N 79.19 deg E]...

Would it make sense to say "Therefore at 25.00s the fish is moving in the direction of this vector:

V1 = 112.0m/s [N 79.19 deg E]"
 
  • #17
gneill
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Your final position vector looks okay.
The final velocity (your V1) seems a bit high to me. How did you calculate it?
 

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