Hey! Sign in to get help with your study questionsNew here? Join for free to post
 You are Here: Home >< Maths

# m1 help

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

question 6d

So far i've gotten t= but i've subbed this into the formula for the position vector of s at time t to get 7i + 117/8 j
help not sure where to go from here

Edit:also i need help with this question too
Question 7 d
I drew my diagram with acceleration now going left instead of right because the question says that the system decelerates but in the answer the arrow for acceleration goes right why is this?

https://2802a3b1a650824d2586fd3336bd...%20Edexcel.pdf
Edit2:question 6 c
i'm managed to draw a diagram and deduce that time 2 position vectors are -2i at 10:00 and -5i+4j at 10:30 however although my diagram tells me that the j component of L is 0 i don't know how to find the i component.

Edit3: question 7c ii no idea where to start
2. (Original post by Big white)
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

question 6d

So far i've gotten t= but i've subbed this into the formula for the position vector of s at time t to get 7i + 117/8 j
help not sure where to go from here

Edit:also i need help with this question too
Question 7 d
I drew my diagram with acceleration now going left instead of right because the question says that the system decelerates but in the answer the arrow for acceleration goes right why is this?

https://2802a3b1a650824d2586fd3336bd...%20Edexcel.pdf
Edit2:question 6 c
i'm managed to draw a diagram and deduce that time 2 position vectors are -2i at 10:00 and -5i+4j at 10:30 however although my diagram tells me that the j component of L is 0 i don't know how to find the i component.

Edit3: question 7c ii no idea where to start
In response to your questions.

(Original post by Big white)
question 6d So far i've gotten t= but i've subbed this into the formula for the position vector of s at time t to get 7i + 117/8 j help not sure where to go from here
This is absolutely correct. Therefore the distance of S from the lighthouse is km as they have the same i component.

(Original post by Big white)
also i need help with this question tooQuestion 7 dI drew my diagram with acceleration now going left instead of right because the question says that the system decelerates but in the answer the arrow for acceleration goes right why is this?
You are correct that it goes to the left. This is negative acceleration and the mark scheme has negative acceleration with the forces acting to the left.

(Original post by Big white)
Edit2:question 6 ci'm managed to draw a diagram and deduce that time 2 position vectors are -2i at 10:00 and -5i+4j at 10:30 however although my diagram tells me that the j component of L is 0 i don't know how to find the i component.
The j component of the lighthouse is 0. Now consider this question. If the lighthouse is southwest of the ship it should have a position vector of given that the j component is 0, find the value of k and therefore find the position vector of the lighthouse.

(Original post by Big white)
question 7c ii no idea where to start
When the velocity is parallel to the velocity must satisfy the equation use the same principle descried above by splitting into the respective i and j components to set up 2 simultaneous equations.
3. (Original post by Cryptokyo)
In response to your questions.

This is absolutely correct. Therefore the distance of S from the lighthouse is km as they have the same i component.

You are correct that it goes to the left. This is negative acceleration and the mark scheme has negative acceleration with the forces acting to the left.

The j component of the lighthouse is 0. Now consider this question. If the lighthouse is southwest of the ship it should have a position vector of given that the j component is 0, find the value of k and therefore find the position vector of the lighthouse.

When the velocity is parallel to the velocity must satisfy the equation use the same principle descried above by splitting into the respective i and j components to set up 2 simultaneous equations.
I see thanks, i didn't see that.

Since the acceleration goes to the left and the resistive forces also goes to the left, how can there be a difference in signs? which ever way you define positive in the signs will be the same for both sides, yet i know that deceleration is negative acceleration, so where does the negative sign come in? Edit: <---- i know how silly that sounds, i got it now (this question anyway)

Where does this come from??? How did you make it? Is there a rule which allows you to construct that??

Again, i don't know where you've gotten that equation from also how am i supposed to solve it??? i can't see any way to do it.
4. (Original post by Big white)

Where does this come from??? How did you make it? Is there a rule which allows you to construct that??

Again, i don't know where you've gotten that equation from also how am i supposed to solve it??? i can't see any way to do it.
When a vector is parallel to the vector it can differ in magnitude hence the vector must be equal to where is a constant term.
For the example of 7 c i on the M1 silver paper, the velocity is parallel to the vector hence the vector must be equal to so .
By considering the i components of the vector only you obtain so .
I know you can already do that bit but it is important to understand.

For 7 c ii the velocity, , is parallel to the vector so we say that
By considering the i components only you get the equation and by considering only the j components you get the equation .
By subbing the first equation into the other you get hence so
5. (Original post by Cryptokyo)
When a vector is parallel to the vector it can differ in magnitude hence the vector must be equal to where is a constant term.
For the example of 7 c i on the M1 silver paper, the velocity is parallel to the vector hence the vector must be equal to so .
By considering the i components of the vector only you obtain so .
I know you can already do that bit but it is important to understand.

For 7 c ii the velocity, , is parallel to the vector so we say that
By considering the i components only you get the equation and by considering only the j components you get the equation .
By subbing the first equation into the other you get hence so
Right i just put it equal to 0 since there was no j component(since it was 0)

Ok thanks i understand now

Write a reply…

Submit reply

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
that username has been taken, please choose another Forgotten your password?
2. this can't be left blank
this email is already registered. Forgotten your password?
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
your full birthday is required
1. Oops, you need to agree to our Ts&Cs to register
2. Slide to join now Processing…

Updated: June 6, 2016
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:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read here first

### 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
Study resources

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

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