m1 help

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question 6d

So far i've gotten t=$2\dfrac{3}{4}$ 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?

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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

question 6d So far i've gotten t=$2\dfrac{3}{4}$ 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

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?

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.

question 7c ii no idea where to start

In response to your questions.



This is absolutely correct. Therefore the distance of S from the lighthouse is $\frac{117}{8}-12.5=2.125$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 $r=-5i+4j+k(-i-j)$ 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 $-i-3j$ the velocity must satisfy the equation $(1-2t)i+(3t-3)j=k(-i-3j)$ use the same principle descried above by splitting into the respective i and j components to set up 2 simultaneous equations.

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 $xi+yj$ it can differ in magnitude hence the vector must be equal to $k(xi+yj)$ where $k$ is a constant term.
For the example of 7 c i on the M1 silver paper, the velocity is parallel to the vector $0i+j$ hence the vector $v=(1-2t)i+(3t-3)j$ must be equal to $k(0i+j)$ so $(1-2t)i+(3t-3)j=0i+kj$.
By considering the i components of the vector only you obtain $1-2t=0$ so $t=0.5$.
I know you can already do that bit but it is important to understand.

For 7 c ii the velocity, $v=(1-2t)i+(3t-3)j$, is parallel to the vector $-i-3j$ so we say that $(1-2t)i+(3t-3)j=k(-i-3j)$
By considering the i components only you get the equation $1-2t=-k$ and by considering only the j components you get the equation $3t-3=-3k$.
By subbing the first equation into the other you get $3t-3=3(1-2t)$ hence $9t=6$ so $t=\frac{2}{3}$