Could someone help me with this maths mechanics

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

(edited 7 years ago)

Reply 1

username2769500

3

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)

Reply 2

username2769500

3

The angle is a(dot)b/|a|x|b| =cos(angle)

Reply 3

Mf1999

OP

11

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

Reply 4

Notnek

21

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.

Reply 5

solC

14

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?

Reply 6

solC

14

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

(edited 7 years ago)

Reply 7

Mf1999

OP

11

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

Reply 8

Mf1999

OP

11

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

Reply 9

Notnek

21

Original post by Mf1999

That's strange it's part of my m1 homework but thank u anyway

Which exam board?

Reply 10

Mf1999

OP

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

Reply 11

Mf1999

OP

11

Original post by notnek

Which exam board?

edexcel

Reply 12

username2769500

3

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.

Reply 13

solC

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

tan\theta = \frac{y}{x}

where y and x are the J and i components of the velocity respectively.

Were you taught a different way in class?

Reply 14

The_Big_E

11

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:

\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

Unparseable latex formula:

\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

PART B:

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.

(edited 7 years ago)

Could someone help me with this maths mechanics

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you