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xxAmyxx
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#1
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Can someone do part C for me please? Thanks.
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MRLX69
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#2
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of course, 1 min....... the Vampire Mathmo is at work.. Muah ha ha!!!!!
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MRLX69
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#3
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Bearing of B from A.... what exactly does that mean? I have an idea but it might be wrong....
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MRLX69
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#4
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Do this:
Find the time when the two i's from A and B are the same. If the i's are the same then they are on top of each other. Therefore the bearing of B from A is 045
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xxAmyxx
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#5
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If they are on top of each other, why is the bearing 45 degrees?
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Christophicus
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For part c.

the relative vector (300-10t)i + 10tj must be parallel to the vector i + j
E.g. tan 1/1 = 45

Therefore;

1/(300-10t) = 1/(10t) = a constant

Cross multiply giving.

300 - 10t = 10t
20t = 300
t = 300/20 = 15 seconds
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Christophicus
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For part d

The magnitude of the vector AB ---> = 300 (for the cars to be 300m apart)

Therefore;
square root ((300-10t)^2 + (10t)^2) = 300

Square both sides

90000 - 6000t + 200t^2 = 90000
200t^2 - 6000t + 90000 = 90000
200t^2 - 6000t = 0
t(200t - 6000) = 0

t = 0 (stated in the question) or t = 6000/200 = 30 seconds
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xxAmyxx
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(Original post by Widowmaker)
For part c.

the relative vector (300-10t)i + 10tj must be parallel to the vector i + j
E.g. tan 1/1 = 45

Therefore;

1/(300-10t) = 1/(10t) = a constant

Cross multiply giving.

300 - 10t = 10t
20t = 300
t = 300/20 = 15 seconds
Sorry for being thick, but what are you doing there? Are you dividing the co-eficcient of i + j by the coefficents of the relative vector?
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Christophicus
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(Original post by xxAmyxx)
Sorry for being thick, but what are you doing there?
If two vectors are parallel.
E.g. ai + bj = ci + dj

c/a = d/b = a constant

For example

2i + 2j is parallel to i + j

so i + j = k(i + j) where k is a constant and in this case it's two.


I've just done the same for the example.

(300-10t)i + 10tj // i + j
therefore;

using the principle c/a = d/b = a constant

1/(300-10t) = 1/(10t) = a constant

Then cross-multiply.

Or you could say that;

10t/(300-10t) = 1

tan^-1 1 = 45 degrees

10t = 300-10t

etc
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Christophicus
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#10
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Do you have the answers to those questions?
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Gauss
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#11
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(Original post by xxAmyxx)
Sorry for being thick, but what are you doing there? Are you dividing the co-eficcient of i + j by the coefficents of the relative vector?
hes just making the gradient equal to 1, so that tan(theta)=45.

the gradient is therefore,

10t/(300-10t) = 1

which is what he did, i think.
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Christophicus
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(Original post by Euclid)
hes just making the gradient equal to 1, so that tan(theta)=45.

the gradient is therefore,

10t/(300-10t) = 1

which is what he did, i think.
yeah that's another way.

Parallel vectors is a very discrete section in M1.
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MRLX69
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#13
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Sorry for not being much help... I was too lazy to do the question
Sorry!
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