Hey there! Sign in to join this conversationNew here? Join for free

Could someone help me with this maths mechanics Watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    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
    Offline

    3
    ReputationRep:
    (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)
    Offline

    3
    ReputationRep:
    The angle is a(dot)b/|a|x|b| =cos(angle)
    • Thread Starter
    Offline

    10
    ReputationRep:
    (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
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (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.
    Offline

    11
    ReputationRep:
    (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?
    Offline

    11
    ReputationRep:
    (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
    Name:  ECTOr.JPG
Views: 35
Size:  5.5 KB
    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
    • Thread Starter
    Offline

    10
    ReputationRep:
    (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
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by solC)
    For part A), you can simply use Velocity = displacement/time
    For part B), A quick diagram may help
    Name:  ECTOr.JPG
Views: 35
Size:  5.5 KB
    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
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by Mf1999)
    That's strange it's part of my m1 homework but thank u anyway
    Which exam board?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (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
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by notnek)
    Which exam board?
    edexcel
    Offline

    3
    ReputationRep:
    (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.
    Offline

    11
    ReputationRep:
    (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 tan\theta = \frac{y}{x} where y and x are the J and i components of the velocity respectively.
    Name:  ECTOr.JPG
Views: 30
Size:  7.2 KB


    Were you taught a different way in class?
    Offline

    9
    ReputationRep:
    (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
    Name:  separator.png
Views: 29
Size:  132 Bytes
    PART A:
    Name:  separator.png
Views: 29
Size:  132 Bytes

    Note that these are position vectors, so they are relative to the origin. This means that:

    \vec{a} = \vec{OA} = \begin{pmatrix}2 \\ -1\end{pmatrix} \ \ \ \ \vec{b} = \vec{OB} = \begin{pmatrix}6 \\ 1\end{pmatrix}

    I've just called the vectors \vec{a} and \vec{b} to make it simpler.

    We are told that the particle is initially at point A and 2s later it is at B. The first thing we need to do is calculate the displacement from A to B. This is denoted by the vector \vec{AB}. We will call this s. With position vectors:

    \vec{s} = \vec{AB} = \vec{b} - \vec{a}

    If you don't understand this step, try using a diagram.

    This gives us:

    \vec{s} = \begin{pmatrix}6 \\ 1\end{pmatrix} - \begin{pmatrix}2 \\ -1\end{pmatrix} = \begin{pmatrix}4 \\ 2\end{pmatrix}

    As \vec{v} = \displaystyle \frac{\displaystyle \vec{s}}{\displaystyle t}}, substituting t = 2 and \vec{s} that we just calculated:

    \vec{v} = \begin{pmatrix}4 \\ 2\end{pmatrix} \div 2

    \vec{v} = \begin{pmatrix}2 \\ 1\end{pmatrix} = 2i + j

    Name:  separator.png
Views: 29
Size:  132 Bytes
    PART B:
    Name:  separator.png
Views: 29
Size:  132 Bytes

    So for part B, we can take \vec{v} and put it on a diagram, starting from the origin like a position vector. The unit vector \vec{i} is parallel to the x_+ axis. Essentially, we need to find the angle \theta between the vector \vec{v} and the x_+ axis. We know that the horizontal component of \vec{v} is 2 and the vertical component is 1. Drawing a triangle, we can see that this is basic trigonometry which you should be able to solve from there.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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