[Pomponz]

Hey guys
Like everyone, I am really confused if the answer for 7c is 1.41 or 5.83 ms-1.
Therefore I thought I would look at the Aqa book, Advancing maths for AQA 2nd edition Mechanics 1.
In this book, in the chapter 3 page 44 there's a worked example that is very similar to the question asked on the exam.
I'll copy it.

An aeroplane has a constant velocity of 120i ms-1 , as it is moving along a runway. It then experiences an accelaration of (2i+5j)ms-2 for the first 20 seconds of its flight. The unit vectors i and j are directed horizontally and vertically, respectively. Assume that the aeroplane is at the origin when it begins to accelerate.

a) Find an expression for the position of the aeroplane at time t seconds, after it starts to accelerate.

r=ut+0.5at^2

r=120ti + 0.5(2i+5j)t^2

r=(120t +t^2)i + (5/2)t^2j



b)Find the speed of the aeroplane when it is at a height of 250m. (This is the similar question to 7c.)


The height of the aeroplane is given by the vertical or j component of the position vector. When the height of the aeroplane is 250 me have:

(5/2)t^2=250
t^2=100
t=10s

The velocity can now be found using this value for t.

v=u+at
v=120i + (2i+5j)x10
v=140i + 50j

The speed can be found as it will be the magnitude if the velocity

v^2= (140)^2 + (50)^2
v= 149 msn-1


PS: Sorry for the way I wrote things. I am not very good with computers.
I hope this helps with deciding which answer is right!
This has been killing me for days! >.<

Thanks

[x_Raman_96]

(Original post by Pomponz)
Hey guys
Like everyone, I am really confused if the answer for 7c is 1.41 or 5.83 ms-1.
Therefore I thought I would look at the Aqa book, Advancing maths for AQA 2nd edition Mechanics 1.
In this book, in the chapter 3 page 44 there's a worked example that is very similar to the question asked on the exam.
I'll copy it.

An aeroplane has a constant velocity of 120i ms-1 , as it is moving along a runway. It then experiences an accelaration of (2i+5j)ms-2 for the first 20 seconds of its flight. The unit vectors i and j are directed horizontally and vertically, respectively. Assume that the aeroplane is at the origin when it begins to accelerate.

a) Find an expression for the position of the aeroplane at time t seconds, after it starts to accelerate.

r=ut+0.5at^2

r=120ti + 0.5(2i+5j)t^2

r=(120t +t^2)i + (5/2)t^2j



b)Find the speed of the aeroplane when it is at a height of 250m. (This is the similar question to 7c.)


The height of the aeroplane is given by the vertical or j component of the position vector. When the height of the aeroplane is 250 me have:

(5/2)t^2=250
t^2=100
t=10s

The velocity can now be found using this value for t.

v=u+at
v=120i + (2i+5j)x10
v=140i + 50j

The speed can be found as it will be the magnitude if the velocity

v^2= (140)^2 + (50)^2
v= 149 msn-1


PS: Sorry for the way I wrote things. I am not very good with computers.
I hope this helps with deciding which answer is right!
This has been killing me for days! >.<

Thanks

Have you got a copy of the paper

[x_Raman_96]

(Original post by Pomponz)
Hey guys
Like everyone, I am really confused if the answer for 7c is 1.41 or 5.83 ms-1.
Therefore I thought I would look at the Aqa book, Advancing maths for AQA 2nd edition Mechanics 1.
In this book, in the chapter 3 page 44 there's a worked example that is very similar to the question asked on the exam.
I'll copy it.

An aeroplane has a constant velocity of 120i ms-1 , as it is moving along a runway. It then experiences an accelaration of (2i+5j)ms-2 for the first 20 seconds of its flight. The unit vectors i and j are directed horizontally and vertically, respectively. Assume that the aeroplane is at the origin when it begins to accelerate.

a) Find an expression for the position of the aeroplane at time t seconds, after it starts to accelerate.

r=ut+0.5at^2

r=120ti + 0.5(2i+5j)t^2

r=(120t +t^2)i + (5/2)t^2j



b)Find the speed of the aeroplane when it is at a height of 250m. (This is the similar question to 7c.)


The height of the aeroplane is given by the vertical or j component of the position vector. When the height of the aeroplane is 250 me have:

(5/2)t^2=250
t^2=100
t=10s

The velocity can now be found using this value for t.

v=u+at
v=120i + (2i+5j)x10
v=140i + 50j

The speed can be found as it will be the magnitude if the velocity

v^2= (140)^2 + (50)^2
v= 149 msn-1


PS: Sorry for the way I wrote things. I am not very good with computers.
I hope this helps with deciding which answer is right!
This has been killing me for days! >.<

Thanks

I followed this exact method, but I multiplied j by -1 and did i = -j on the position vector to get south east. Then I inserted the time in to v = u + at. Then I used pyhtag, with 3i + -5j and got 5.83

[Pomponz]

(Original post by x_Raman_96)
Have you got a copy of the paper

No, not yet but I know the question was very similar to this.
But you never know, I am human, so I can always be wrong.

[Pomponz]

(Original post by x_Raman_96)
I followed this exact method, but I multiplied j by -1 and did i = -j on the position vector to get south east. Then I inserted the time in to v = u + at. Then I used pyhtag, with 3i + -5j and got 5.83

Yes I did the same as they asked for SOUTHEAST
SO It should have been i=-j like you say.
I was just wondering if this was the right way of solving the problem thou.
For the looks of what the book says it is but I would like to see the exact question again to be 100%.

[Pomponz]

Ok I found another exercise which supports the followers of 1.41 ms-1
Here is the question, from the Revision mechanics book from aqa.


A particle moves with constant velocity, so that during a 10 second period of time its velocity changes from (9i-2j)ms-1 to (-i +23j)ms-1. The unit vectors i and j are directed east and north respectively.


a)Find the acceleration of the particle.

-i+23j=9i-2j + 10a

a=-i+2.5j

b)Find the speed of the particle when it is moving due east Similar question to the exam one (7c).

v=(9i-2j) + (-i+2.5j)t

v=(9-t)i + (-2+ 2.5t)j


When the particle is travelling due east, the j component of the velocity will be zero.

-2+2.5t=0

t=(2/2.5)

t=0.8

The velocity at this time can now be found.

v(0.8)=(9-0.8)i= 8.2i

The speed of the particle will be 8.2 ms-1.


This is a worked example from page 13 of "Revise for Mechanics 1"


So like I said, one word could change the way the exercise would be done, so only when I see the actual question I can see which one is right.
Although I have a weird sence on my gut this may be the right answer!

Thanks

[Jit-Mistry]

(Original post by x_Raman_96)
I followed this exact method, but I multiplied j by -1 and did i = -j on the position vector to get south east. Then I inserted the time in to v = u + at. Then I used pyhtag, with 3i + -5j and got 5.83

your wrong/.... if something is travelling southeast the i and j are the same number except j is negative one...

[Pomponz]

(Original post by Jit-Mistry)
your wrong/.... if something is travelling southeast the i and j are the same number except j is negative one...

Yes it becomes i=-j

[karchun]

Okay lets just end this Q's 7 feud.... the answer is 1.41, √2... a bunch of my class got this score... even my 2 classmates who score 100 on previous m1 papers...

Moving on; Anyone got the paper cause I really want to know what marks each question was!

This was posted from The Student Room's Android App on my GT-N7000

[x_Raman_96]

****ers.

[karchun]

(Original post by x_Raman_96)
****ers.

This was posted from The Student Room's Android App on my GT-N7000

[x_Raman_96]

(Original post by Jit-Mistry)
your wrong/.... if something is travelling southeast the i and j are the same number except j is negative one...

you're telling me 1 = -1. that's the most *******s I've heard 3 doesn't = -3. x does not = -x.

[karchun]

(Original post by x_Raman_96)
you're telling me 1 = -1. that's the most *******s I've heard 3 doesn't = -3. x does not = -x.

In this respect I think he means the theory of I and j being the same; use of Pythagoras theorem would assume I= J - this is only an assumption for when I and J are positive so when the direction s east is involved it would mean j has turned to negative hence I=-J ...

This was posted from The Student Room's Android App on my GT-N7000

[Oromis263]

(Original post by x_Raman_96)
you're telling me 1 = -1. that's the most *******s I've heard 3 doesn't = -3. x does not = -x.

They are vectors, but what he means is the magnitude of both of the results must be the same if an object is has a resultant velocity in a south-eastern direction.

[Retro.spex]

(Original post by croasdaj)
The question asked specifically about the block which it never said was treated like a particle

I thought exactly the same. Hope we're right!

[x_Raman_96]

It's 5.83.

[Oromis263]

(Original post by x_Raman_96)
It's 5.83.

Upload the paper and prove it. Until then, I will remain happy with knowing I read the question correctly.

[x_Raman_96]

It is definitely 5.83

[samkanwoods]

(Original post by x_Raman_96)
It is definitely 5.83

I got 5.83, but its is wrong as how would it be travelling south-east if he had a velocity of 3i-5j or 5i-3j (cant remember). So root2 wouls be right as this would give a velocity of of 1i-1j, which means that he is going one to the right and one to the bottom. So he would be going south-east. So it is NOT 5.38 sadly.

[x_Raman_96]

(Original post by samkanwoods)
I got 5.83, but its is wrong as how would it be travelling south-east if he had a velocity of 3i-5j or 5i-3j (cant remember). So root2 wouls be right as this would give a velocity of of 1i-1j, which means that he is going one to the right and one to the bottom. So he would be going south-east. So it is NOT 5.38 sadly.

It is 5.83