SOLUTION: Hey again! :D Im stuck on this question about vectors including acceleration and relative positioning. The question is.. A particle has position 2i + j initially and is movin

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Question 106431This question is from textbook Mechanics I
: Hey again! :D Im stuck on this question about vectors including acceleration and relative positioning.
The question is..
A particle has position 2i + j initially and is moving with speed 10ms-1 in the direction 3i - 4j. Find it's position vector when t = 3 and the distance it has travelled in those 3 seconds.
I know the distance travelled is Speed * time, so 10 * 3 = 30metres. But when i worked out the relative position i got it totally wrong, does anyone know how to do this? Thanks alot in advance! :DThis question is from textbook Mechanics I

Found 3 solutions by Fombitz, stanbon, MathLover1:
Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, 30 m is the magnitude of the position change vector (AB) but the direction is (3,-4) in (i,j) coordinates.
You start at A (2,1) and go to B (2+i component of AB,1+j component of AB).
Find the magnitudes of the i component and the j component using the angle with vertex at A.
drawing%28+300%2C+300%2C+-10%2C+30%2C+-30%2C+10%2Cgrid%28+1+%29%2Cgreen%28circle%28+2%2C+1%2C+.5+%29%29%2Cgreen%28circle%28+20%2C+-23%2C+.5+%29%29%2Cred%28line%282%2C1%2C20%2C-23%29%29%2Clocate%282.5%2C4%2CA%29%2Clocate%2821%2C-23%2CB%29%2Cblue%28line%282%2C1%2C20%2C1%29%29%2Cblue%28line%2820%2C1%2C20%2C-23%29%29%29
Hint : 3,4,5 make up a Pythagorean triple.
Sine and cosine of an angle using these sides are either 3/5 or 4/5.

Answer by stanbon(48535) About Me  (Show Source):
You can put this solution on YOUR website!
A particle has position 2i + j initially and is moving with speed 10ms-1 in the direction 3i - 4j. Find it's position vector when t = 3 and the distance it has travelled in those 3 seconds.
I know the distance travelled is Speed * time, so 10 * 3 = 30metres.
-------------
Draw the figure:
A vector from (2,1) thru (3,-4) and extended to (x,y)
A vertical segment from (2,1) down to (2,-4)
A horizontal segment form (2,-4) to (3,-4)
These form a triangle with hypotenuse = sqrt(26),
vertical = 5, horizontal = 1.
----------
Need to find x and y.
-----------------
The diagonal distance is 30 meters as you pointed out.
So x^2 + y^2 = 900
-----------------------
Draw a vertical line from (2,1) down to (2,y)
Draw a horizontal line from (2,y) to (x,y)
------------------------
You now have two proportional triangles.
Vertical Proportion: 30/sqrt(26) = v/5
v = 150/sqrt26 = 29.417
--------
Horizontal Proportion:
30/sqrt(26)= h/1
h = 5.88
--------------
Therefore:
x = 2+h = 7.88
y = 1-29.417 = -28.417
--------------
These are the position coordinates after 3 seconds.
====================
Cheers,
Stan H.

Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
The speed is 10+ms%5E%28-1%29; let assume there is no acceleration, and the particle travels 30+m in 3+s.
Since we need to find a vector with magnitude 30 and parallel to 3i+-%284j%29, to do this first find a unit vector parallel to 3i+-+%284j%29 and multiply it by 30.
A unit vector parallel to 3i+-+%284j%29 is:
%283i+-+%284j%29%29+%2F%283i+-+%284j%29%29+
%283i+-+%284j%29%29%2Fsqrt%283%5E2+%2B+%28-4%292%29+
%283i+-+%284j%29%29%2Fsqrt%2825%29
%283i+-+%284j%29%29%2F%285%29

So the vector we want is
30%283i+-+%284j%29%29%2F%285%29
=6%283i+-+%284j%29%29
= %2818i+-+%2824j%29%29
Now, add that to the particle’s initial position:
%2818i+-+%2824j%29%2B+2i+%2B+j%29=+%2820i+-+%2823j%29%29
That’s its position vector after 3+seconds