Pure yr2 vectors help needed please

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Sidd1
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I've been stuck on these questions for days now I posted the first question, being Q6, on TSR but sadly haven't gotten the help I needed. I would really appreciate if you could help me out with these two! Sir Cumference

Thanks in advance
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Plücker
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(Original post by Sidd1)
I've been stuck on these questions for days now I posted the first question, being Q6, on TSR but sadly haven't gotten the help I needed. I would really appreciate if you could help me out with these two! sircumference

Thanks in advance
Do you have any working that you can post so that we can see how far you have got and where you might have gone wrong?
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Sidd1
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(Original post by Plücker)
Do you have any working that you can post so that we can see how far you have got and where you might have gone wrong?
Um yes I do for one of the questions only for the other one I didn’t really know where to start 😕
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I’m not sure why the first picture didn’t come out portrait but I hope that’s okay

For the first question (Q6) this is the only thing I did:

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mqb2766
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(Original post by Sidd1)
Um yes I do for one of the questions only for the other one I didn’t really know where to start 😕
Name:  657EC326-7232-4E55-AE48-423241B6CDF5.jpeg
Views: 15
Size:  280.1 KB
Name:  83A4C684-40AA-4C8F-8DED-F0A5DA700364.jpeg
Views: 13
Size:  106.7 KB

I’m not sure why the first picture didn’t come out portrait but I hope that’s okay

For the first question (Q6) this is the only thing I did:

Name:  C5B728E3-2D1D-4CFC-8273-A5939A807C55.jpeg
Views: 14
Size:  132.3 KB
For Q6 b) you need to workout the (sign of) acceleration in the k direction as this is vertical height. What do you get?

Edit - you've worked out an angle, so what problem are you having?
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Sidd1
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(Original post by mqb2766)
For Q6 b) you need to workout the (sign of) acceleration in the k direction as this is vertical height. What do you get?

Edit - you've worked out an angle, so what problem are you having?
The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??
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Plücker
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(Original post by Sidd1)
The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??
k is up.

-460k is down.
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mqb2766
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(Original post by Sidd1)
The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??
It's flying level. What is the resultant force and hence acceleration in the k direction? Positive is up, negative downwards.
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Sidd1
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(Original post by Plücker)
k is up.

-460k is down.
I don't understand how it's -460 because that's part of the resultant force in the k direction. Wouldn't it be -23/60 after working out acceleration using F=ma?
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(Original post by Sidd1)
I don't understand how it's -460 because that's part of the resultant force in the k direction. Wouldn't it be -23/60 after working out acceleration using F=ma?
Once you have the resultant force you don't need any further calculation. The acceleration is proportional to the resultant force and that has a downward component so the acceleration has a downward component. Since the aeroplane is initially flying level and the acceleration has a downward component the aeroplane is descending.
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(Original post by Plücker)
Once you have the resultant force you don't need any further calculation. The acceleration is proportional to the resultant force and that has a downward component so the acceleration has a downward component. Since the aeroplane is initially flying level and the acceleration has a downward component the aeroplane is descending.
Okay so the acceleration being proportional to the resultant force is this because it's in level flight so all the forces are balanced? Force up = force down??
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mqb2766
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(Original post by Sidd1)
Okay so the acceleration being proportional to the resultant force is this because it's in level flight so all the forces are balanced? Force up = force down??
No, the plane has zero vertical velocity at time 0. But there is a force (acceleration) on it.

Imagine a parabola with a maximum at time 0. The derivative is zero, but the second derivative (acceleration) is negative. This causes the curve to decrease.

Or else use
v = u + at
u = 0, a is negative so what will v be as time increases.
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Sidd1
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(Original post by Plücker)
Once you have the resultant force you don't need any further calculation. The acceleration is proportional to the resultant force and that has a downward component so the acceleration has a downward component. Since the aeroplane is initially flying level and the acceleration has a downward component the aeroplane is descending.
Is this what you're talking about? (Attached)
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Sidd1
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(Original post by mqb2766)
No, the plane has zero vertical velocity at time 0. But there is a force (acceleration) on it.

Imagine a parabola with a maximum at time 0. The derivative is zero, but the second derivative (acceleration) is negative. This causes the curve to decrease.

Or else use
v = u + at
u = 0, a is negative so what will v be as time increases.
Oh okay I get the parabola example.
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mqb2766
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(Original post by Sidd1)
Oh okay I get the parabola example.
suvat is just quadratic (parabola) in each dimension for displacement againat time. Acceleration (or force) is the curvature (2nd derivative) term.
The velocity is then the gradient, which is linear (against time).
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(Original post by mqb2766)
suvat is just quadratic (parabola) in each dimension for displacement againat time. Acceleration (or force) is the curvature (2nd derivative) term.
The velocity is then the gradient, which is linear (against time).
I feel like you're making it more complicated than it is for me, no offence😅.

All I want to know is why the resultant force k vector (-460) is the same as the acceleration.
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mqb2766
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(Original post by Sidd1)
I feel like you're making it more complicated than it is for me, no offence😅.
Not at all, if you understand your quadratic curves etc, it can help with suvat, such as finding stattionary points etc.
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mqb2766
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(Original post by Sidd1)
I feel like you're making it more complicated than it is for me, no offence😅.

All I want to know is why the resultant force k vector (-460) is the same as the acceleration.
f = ma
so the force is proportional to acceleration - same sign, but scaled by the mass.
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Plücker
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(Original post by Sidd1)
Is this what you're talking about? (Attached)
Yes, Newton's second law, F = ma, tells us that (for a constant mass) force and acceleration are directly proportional.
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Sidd1
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Also, for the other Question (Q9), I'm struggling how to work out the coordinates of D. I first tried plotting the points in a 3D plane but I struggled:/ I'm not really sure how else to tackle that. When it says A will be reflected in the line BC how do I know which coordinate will change?
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mqb2766
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D will lie in the plane containing A,B,C.
Can you sketch what the shape will look like in that plane? A reflection must mean that AD is perpendicular to BC and D is the same distance the other side.
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