A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2ij)m relative to a fixed origin O and 2s later it is at the point b with position vector (6i+j)m
A) find the velocity of p
B) find the size of the angle between the direction of motion of p and the vector i in degrees
3s after it passes b the particle p reaches the point c
C) find in m the distance oc
Step by step pls

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 05012017 16:56
Last edited by Mf1999; 05012017 at 17:20. 
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 05012017 17:03
(Original post by Mf1999)
A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2ij)m relative to a fixed origin O and 2s later it is at the point b with position vector (6i+j)m
A) find the velocity of p
B) find the size of the angle between the direction of motion of p and the vector i in degrees
3s after it passes b the particle p reaches the point c
C) find in m the distance oc 
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 05012017 17:04
The angle is a(dot)b/axb =cos(angle)

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 05012017 17:20
(Original post by Anfanny)
The angle is a(dot)b/axb =cos(angle) 
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 05012017 17:44
(Original post by Anfanny)
The two vectors différence is the distance. If that is a scalar product of either position vectors It is travelling on one direction And is said to be paralell to it's velocity. Velocity It per second. Or distance in 2s = 1/2(b) or 2(a) where b = 2i j And b = 4i +2j If that is the case You cant work out all the above If magnitude is equal to the Root of the sum of the squares of the modolus of the vectors. Is ROOT(ai^2 +bi^2+aj^2+bj^2)
I don't have time to help now but hopefully someone else can. 
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 05012017 17:47
(Original post by Mf1999)
A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2ij)m relative to a fixed origin O and 2s later it is at the point b with position vector (6i+j)m
A) find the velocity of p
B) find the size of the angle between the direction of motion of p and the vector i in degrees
3s after it passes b the particle p reaches the point c
C) find in m the distance oc
Step by step pls 
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 05012017 17:56
(Original post by solC)
Which part don't you get?(Original post by Mf1999)
A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2ij)m relative to a fixed origin O and 2s later it is at the point b with position vector (6i+j)m
A) find the velocity of p
B) find the size of the angle between the direction of motion of p and the vector i in degrees
3s after it passes b the particle p reaches the point c
C) find in m the distance oc
Step by step pls
For part B), A quick diagram may help
Now you should be able to use right angled triangles to find theta
For part C) You first need to find the displacement 3 seconds after p passes b(using the same equation as in A), and then note that the distance is the modulus of the displacement vectorLast edited by solC; 05012017 at 18:01. 
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 05012017 17:58
(Original post by notnek)
The scalar product is not part of the M1 module.
I don't have time to help now but hopefully someone else can. 
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 05012017 17:59
(Original post by solC)
For part A), you can simply use Velocity = displacement/time
For part B), A quick diagram may help
Now you should be able to use right angled triangles to find theta
For part C) You first need to find the displacement 3 seconds after p passes b(using the same eqution as in A), and then note that the distance is the modulus of the displacement vector 
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 05012017 18:00
(Original post by Mf1999)
That's strange it's part of my m1 homework but thank u anyway 
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 05012017 18:13
(Original post by solC)
Which part don't you get? 
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 05012017 18:14
(Original post by notnek)
Which exam board? 
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 05012017 18:20
(Original post by Mf1999)
How did you do it I don't get it all 
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 05012017 18:24
(Original post by Mf1999)
I'm still kind of unsure of part b we haven't been taught how to use a diagram could you possiably explain it to me
In this case, you would use where y and x are the J and i components of the velocity respectively.
Were you taught a different way in class? 
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 05012017 18:24
(Original post by Mf1999)
A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2ij)m relative to a fixed origin O and 2s later it is at the point b with position vector (6i+j)m
A) find the velocity of p
B) find the size of the angle between the direction of motion of p and the vector i in degrees
3s after it passes b the particle p reaches the point c
C) find in m the distance oc
Step by step pls
PART A:
Note that these are position vectors, so they are relative to the origin. This means that:
I've just called the vectors and to make it simpler.
We are told that the particle is initially at point and later it is at . The first thing we need to do is calculate the displacement from to . This is denoted by the vector . We will call this . With position vectors:
If you don't understand this step, try using a diagram.
This gives us:
As , substituting and that we just calculated:
PART B:
So for part B, we can take and put it on a diagram, starting from the origin like a position vector. The unit vector is parallel to the axis. Essentially, we need to find the angle between the vector and the axis. We know that the horizontal component of is and the vertical component is . Drawing a triangle, we can see that this is basic trigonometry which you should be able to solve from there.Last edited by The_Big_E; 05012017 at 18:28.
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