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! :D
This question is from textbook Mechanics I
Found 3 solutions by Fombitz, stanbon, MathLover1:
Answer by Fombitz(32388) (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.
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(75887) (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(20849) (Show Source): You can put this solution on YOUR website!
Given:
The speed is ; let assume there is no acceleration, and the particle travels in .
Since we need to find a vector with magnitude and parallel to , to do this first find a unit vector parallel to and multiply it by .
A unit vector parallel to is:
So the vector we want is
=
=
Now, add that to the particle’s initial position:
That’s its position vector after seconds
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