# Pure yr2 vectors help needed please

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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

Thanks in advance

Last edited by Sidd1; 8 months ago

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#2

(Original post by

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

**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

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(Original post by

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?

**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?

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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#4

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Um yes I do for one of the questions only for the other one I didn’t really know where to start 😕

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:

**Sidd1**)Um yes I do for one of the questions only for the other one I didn’t really know where to start 😕

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:

Edit - you've worked out an angle, so what problem are you having?

Last edited by mqb2766; 8 months ago

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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?

**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?

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#6

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The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??

**Sidd1**)The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??

**k**is up.

-460

**k**is down.

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#7

**Sidd1**)

The part where it asks whether the aeroplane is ascending or descending. I don’t get why it’s descending ??

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#9

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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?

**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?

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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.

**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.

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#11

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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??

**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??

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.

Last edited by mqb2766; 8 months ago

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**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.

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(Original post by

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.

**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.

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#14

(Original post by

Oh okay I get the parabola example.

**Sidd1**)Oh okay I get the parabola example.

The velocity is then the gradient, which is linear (against time).

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(Original post by

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).

**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).

All I want to know is why the resultant force k vector (-460) is the same as the acceleration.

Last edited by Sidd1; 8 months ago

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#16

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I feel like you're making it more complicated than it is for me, no offence😅.

**Sidd1**)I feel like you're making it more complicated than it is for me, no offence😅.

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#17

(Original post by

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.

**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.

so the force is proportional to acceleration - same sign, but scaled by the mass.

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#18

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Is this what you're talking about? (Attached)

**Sidd1**)Is this what you're talking about? (Attached)

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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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#20

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.

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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