Turn on thread page Beta
 You are Here: Home >< Maths

# Could someone help me with this maths mechanics watch

1. A particle p moves in a straight line with constant velocity. Initially p is a the point a with position vector (2i-j)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
2. (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 (2i-j)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
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)
3. The angle is a(dot)b/|a|x|b| =cos(angle)
4. (Original post by Anfanny)
The angle is a(dot)b/|a|x|b| =cos(angle)
How did you do it I don't get it all
5. (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)
The scalar product is not part of the M1 module.

I don't have time to help now but hopefully someone else can.
6. (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 (2i-j)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
Which part don't you get?
7. (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 (2i-j)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 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 equation as in A), and then note that the distance is the modulus of the displacement vector
8. (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.
That's strange it's part of my m1 homework but thank u anyway
9. (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
Thank u so much
10. (Original post by Mf1999)
That's strange it's part of my m1 homework but thank u anyway
Which exam board?
11. (Original post by solC)
Which part don't you get?
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
12. (Original post by notnek)
Which exam board?
edexcel
13. (Original post by Mf1999)
How did you do it I don't get it all
Dit product is multipy perpendiculair vectors in all the arrangements So ij ik ji ki kj jk as the product of paralell is zero. Its only needed in c4. The properly method may involve drawing two big triangles And pythagoras! I And j are equal but in two different directions.
14. (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
If you drop a perpendicular down to the x-axis, then you end up with a right-angled triangle. Therefore you can use SOHCAHTOA to find the angle required.
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?
15. (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 (2i-j)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.

Reply
Submit reply
Turn on thread page Beta

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 5, 2017
Today on TSR

### Exam Jam 2018

Join thousands of students this half term

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.