m1 aqa practice questions vectors

The three questions appear to be as follows:



5. A particle is initial at point O, with position vector sO = (i + 2j) m. It travels with constant velocity (3i + j) ms-1. After 8 s, it reaches point A. A second particle has constant velocity (-4i + 2j) ms-1 and takes 5 s to travel from point A to point B.

Find sA and sB, the position vectors of points A and B.

6. At t = 0, a particle P is at position vector (i + 5j) m, relative to a fixed origin, where i and j are the unit vectors pointed east and north, respectively. P is moving with a constant velocity of (7i - 3j) ms-1. After 4 s, the velocity of P changes to (ai + bj) ms-1. After a further 3.5 s, P reaches position vector (15i) m. Find:

a) the speed of P at t = 0,

b) the bearing of P at t = 0, and

c) the values of a and b.

7. A particle moves on a smooth horizontal plane. It is initially at point P, with position vector (i + j) m relative to a fixed origin. At P, the particle has velocity (3i - j) ms-1. The particle moves with constant acceleration for 10 s until it reaches point Q, with position vector (6i + 11j) m.

a) Find the particle's acceleration.

b) Show that, 6 seconds after the particle leaves P, it is moving parallel to unit vector j.




Please reply showing your thoughts and efforts so far, so we can see where things are bogging down. Thank you! ;)
 
5.
A particle is initial at point
O, with position vector sO = (i + 2j) m. It travels with constant velocity (3i + j) ms-1. After 8 s, it reaches point A. A second particle has constant velocity (-4i + 2j) ms-1 and takes 5 s to travel from point A to point B.

Find sA and sB, the position vectors of points A and B.
------------------------
sa = position vektor + velocity * times
sa = (i+2j) + 8*( 3i+j) = i+2j + 24i+8j = 25i+10j
at the moment we re at the point of position vector
25i+10j so
sb = (25i+10j) +5
(-4i + 2j) = 5i+20j

6. At t = 0, a particle P is at position vector (i + 5j) m, relative to a fixed origin, where i and j are the unit vectors pointed east and north, respectively. P is moving with a constant velocity of (7i - 3j) ms-1. After 4 s, the velocity of P changes to (ai + bj) ms-1. After a further 3.5 s, P reaches position vector (15i) m. Find:
----------------------
a) the speed of P at t = 0,
the speed of p at t = 0 is magnitude of this velocity vector
(7i - 3j) so speed = root of 58 = 7.61 ms-1
b) the bearing of P at t = 0, and

here i ve got problem to imagine what actually the bearing is. is that the angle between direction and north?
in my book i can see the bearing is tan-1(-3/7) so they find the angle from velocity vector
(7i - 3j)
? why not from the position vector? velocity is speed with direction but it doesnt say anything about position so why could i get information about the angle from it?
c) the values of a and b.
now i need to get the position after 4 seconds so
(i+5j)+(7i-3j)*4 = 29i-7j after another period of time p reaches position vector 15i so s = v*t
15-(29i-7j) = v*3.5 which says that v = (-14i+7j)/3.5
7. A particle moves on a smooth horizontal plane. It is initially at point P, with position vector (i + j) m relative to a fixed origin. At P, the particle has velocity (3i - j) ms-1. The particle moves with constant acceleration for 10 s until it reaches point Q, with position vector (6i + 11j) m.

a) Find the particle's acceleration.
i am struggling with setting up a strategy how to solve it.
first what come in my mind is
s =
5i+10j
u
v=
(3i - j) ms-1

a = ai + bj
t = 10s
variables and equation s = vt - 1/2at^2 but there s not enough known variables.
displacement could be the difference between
s = (6i + 11j) - (i + j) = 5i+10j
5i+10j = (3i - j)10 - ( ai + bj)50
5i+10j-30i+10j=-( ai + bj)50
(-25i+20j)/50 = -( ai + bj)
(-25/50)i +(20/50)j =
-( ai + bj)
1.53i-0.4j =( ai + bj)
the exam book says it is -0.5i+0.4. they used different suvat equation ut+1/2at^2

b) Show that, 6 seconds after the particle leaves P, it is moving parallel to unit vector j.

another struggles, to be honest i am lost at the moment.
to fingure out it s moving parallel i need probably to get the position vector after six seconds so
it is (i+j) + (
3i - j)*6 = 19i-5j
and that is not true. they use v = u+at and end up with v =1.4j.

but i would expect to look for position vector and after compare the values with the others position vectors.

honestly i am pretty confused by this part of maths. i don't know how to start solving those problems, i have problem to imagine what is what as well. finally I start and i ll end up with wrong result because i picked up wrong equation. maybe I should just work with it till i get it.

thank you for every advice.

 
vector parallel to the unit j

hello,

question 7. b)

https://onedrive.live.com/redir?res...authkey=!ALLDoVLKCqgGIz8&v=3&ithint=photo,jpg

they re asking to prove that after six seconds the particle is moving parallel to the unit vector J.

https://onedrive.live.com/redir?res...authkey=!AO-f0SpxtNTerJY&v=3&ithint=photo,jpg

on the second picture you can see solution in my test book. I don't understand how the solution prove the fact it s moving along the unit vector J???

Thank you for any advice.
 
hello,

question 7. b)

https://onedrive.live.com/redir?res...authkey=!ALLDoVLKCqgGIz8&v=3&ithint=photo,jpg

they re asking to prove that after six seconds the particle is moving parallel to the unit vector J.

https://onedrive.live.com/redir?res...authkey=!AO-f0SpxtNTerJY&v=3&ithint=photo,jpg

on the second picture you can see solution in my test book. I don't understand how the solution prove the fact it s moving along the unit vector J???

Thank you for any advice.

Did you calculate the part 'a' - acceleration vector with magnitude and direction?

If you did - what did you get? - how did you get it - please share your work.

If you didn't - do that first!
 
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