M1 Vector question- please help!

Two cars A and B are moving on straight horizontal roads with constant velocities. The velocity of
A is 20 ms-1 due east, and the velocity of B is (10i + 10j) ms-1, where i and j are unit vectors
directed due east and due north respectively. Initially A is at the fixed origin O, and the position
vector of B is 300i m relative to O. At time t seconds, the position vectors of A and B are r metres
and s metres respectively.

(a) Find expressions for r and s in terms of t. (3)
(b) Hence write down an expression for AB in terms of t. (1)
(c) Find the time when the bearing of B from A is 045. (5)
(d) Find the time when the cars are again 300 m apart. (6)

I can do a and b but what about c? help, time is running out.......... :eek:

Reply 1

DaKe

sash37uk

Two cars A and B are moving on straight horizontal roads with constant velocities. The velocity of
A is 20 ms-1 due east, and the velocity of B is (10i + 10j) ms-1, where i and j are unit vectors
directed due east and due north respectively. Initially A is at the fixed origin O, and the position
vector of B is 300i m relative to O. At time t seconds, the position vectors of A and B are r metres
and s metres respectively.

(a) Find expressions for r and s in terms of t. (3)
(b) Hence write down an expression for AB in terms of t. (1)
(c) Find the time when the bearing of B from A is 045. (5)
(d) Find the time when the cars are again 300 m apart. (6)

I can do a and b but what about c? help, time is running out.......... :eek:

I think i know what to do but can u post up the answers to the first three parts....

I get the answer to be (hope they are right... :smile:

)

(a) r = 20t, s = 10(2^½)
(b) AB = (10(2^½)-20)t
(c) t = 15s

I can do parts (c) and most of part (d), but I'm not sure how to explain them here, as I use a diagram and can't show you one...

Reply 3

Sp123

You should end up with a quadratic, I know that...

I'm gonna consult the book, see what I can get

Reply 4

DaKe

ok for (b) i got something completely different....

i got...300i + (10i + 10j)t

cos i thought u had to express in terms of t?

I'll try part (c) though...

A travels in a straight line alone the horistonal axis, and always stays on there. B travels away from the horiztonal axis at 45° in a positive direction horizontally and vertically.

With O the origin, where A started from, we can draw a diagram consiting of OA, OB and AB.

We see that AB must form an angle of 45° with the horizontal axis due to the conditions in the question.

But B travelled along such a line, thus we conclude that AB is the line which car B travelled along. Hence at this point in time, A must be in the position from which B started at.

Therefore A has travelled 300m at 20ms^-1. This tells you that the time i t = 15s.

Reply 6

DaKe

your part (c) seems correct t=15

DaKe

ok for (b) i got something completely different....

i got...300i + (10i + 10j)t

cos i thought u had to express in terms of t?

I multiplied out the B vector to remove the i's and j's....here is how I did it...

We want the distance AB in terms of time. So for a general time T we need to find the distance vectors from the origin to A and to B. we worked these out in (a) so let's call the R and S respectively.

Then AB = S-R by deffinition of the distance between two vectors.

Now S= (10i + 10j)t. Multiplying out this vector, using Pythag, we get S^2 = (10t)^2 + (10t)^2 = 200t^2. Therefore S = 10(2^½)t

We had R = 20t.

Sp S-R = 10(2^½)t - 20 t = (10(2^½)-20)t

Reply 8

sash37uk

OP

1

On the mark scheme they had this:

10t/ 300-10t= 1

How did they get this..

sash37uk

On the mark scheme they had this:

10t/ 300-10t= 1

How did they get this..

Which part of the question is that for? Part (c) as there may be a way to do the question using that equation...let me think...

Reply 10

sash37uk

OP

1

yeh it was for c. Is it something to do with AB being// to i+j ?

Reply 11

sash37uk

OP

1

this does give an answer of t=15, but I dont understand what that equation is or does.

Reply 12

sash37uk

OP

1

Ok i think ive got it now, j/i is the gradent which is parallel to AB, therefore 10t/ 300-10t= 1 (because its at 45 degrees. and so i and j will have the same value) thanks for your help neway :wink:

sash37uk

yeh it was for c. Is it something to do with AB being// to i+j ?

Right we know B travels at (10i +10j)ms^-1, thus we see that B has travelled 10t m horizontally, making B 300 +10t alonmg the horizontal axis. Suppose B is directly above the point K on the horizontal axis.

We know A has travelled 20t m horizontally in this time along the horizontal axis. Thus AK = OK - OA = 300 +10t -20t = 300 - 10t.

Now in this time B has travelled 10t m vertically (as B has a velocity of (10i+10j)ms^-1. So BK = 10t

Now consider the triangle ABK. And because AB makes an angle of 45° with the verticle we know that AK = BK in length.

Thus the ratio of these two sides is equal and dividing the lenght of one side with the other gives us 1.

Hence BK/AK = 10t / (300-10t) = 1.

Rearangin this to find t gives you the answer t=15s.

Don't worry if you couldn't have come up with this same answer, as it is only one possible method. I originally came up with a different method to both this one and the one I gave above...there are many ways to do it and unless you are asked for a specific way any correct method should be OK.

M1 Vector question- please help!

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