Question thread

[kamil9876]

4D) I get how the closest distance between two vectors is when they're perpendicular but why is it the velocity and displacement vector? I can't see the "connection" between these two. At first I thought the derivative of the position vector = 0 would give the max/min but obviously that didn't go too far.

Firstly note that the direction of the velocity vector is the direction of the tangent of the curve. So in the case of straight line the direction of the velocity vector is the straight line, I could extend this argument to a concave curve but my paint skillz suck .

Now let the minimum distance be and construct a circle of radius and centre at the origin. The curve(line) will just touch the circle at some point but will not go inside the circle, otherwise it wouldn't be the minimum distance since inside the circle it is closer to the origin than . Therefore the straight line is the tangent of the circle and hence perpendicular to the radius at that point.

[kamil9876]

oh and you were close with the derivative of position vector=0, except you would need to take the derivative of the modulus of position vector and equate it to 0.

[kamil9876]

Q9) A small object is dropped vertically down from the top of a building which is h metres high. It takes 0.4 seconds to travel the last 8 metres before it hits the ground. If the air resistance is negligible, show that the height of the building is:

after travelling a distance of , initial velocity being , final velocity being . This gives:


then in the second part of the trip i has initial velocity and travels for 0.4 seconds:


now find expression for and sub into the first expression to find .

1a) Vectors u = a + kb and v = a - b are perpendicular and |b| = 2|a|. Find the value of k for which the angle between the vectors a and b is 120 degrees.
(I got the 1st equation but I don't get the solutions from then on)

but

lot of manipulations, hopefully no mistakes 

[TonyHem]

http://www.vcaa.vic.edu.au/vce/studies/mathematics/specialist/pastexams/2006/2006specmath.pdf

For exam 2, extended response question 5E ( last question)
I got all the previous parts but ( horizontal distance = 2.1m for reference)
I did something different to the itute solutions and my answer is slightly off.
They break the situation into a horizontal and vertical components and then take the modulus of the velocity(i,j components) * 75 (kg) to get the magnitude of the momentum.
The speed at the end of the ramp is 6.2ms^-1, and I had t = 0.4, v = unknown and distance
So I worked out landing speed ( v) to be = 8.533 and then got P to be 640 kgms^-1
The answers get V to be 8.8, so P is 660.
Thanks for any help!

[kamil9876]

Can you explain in greater detail how you did this:

*t=0.4 was calculated from constant acceleration formula?
*"distance" was calculated from pythagoras' theorem?
*how did you use constant acceleration formulas?: what did you set as initial velocity? etc.

I think the problem lies in a few things: Using constant acceleration formulas with initial velocity of one component but final velocity assumed to be in another component. Also the distance travelled is not calculated from simply pythagoras' theorem since it undergoes projectile motion rather than motion in a striaght line, and besides, I do not know how distance travelled would be helpful.

[TonyHem]

Umm I used initial speed from the previous questions which was 6.2ms^-1 at the end of the slide
I got t = 0.4 from part D, using vertical components (x = 2m, u = 6.2sin(theta) t=? a = g)
then did t = 0.4, x = the above thing, and used x = ((u+v)/2)t
I thought since the person was sliding down like \, I kinda imagined something below, where the tip is the end of the slide, the vertical to be 2m and then the horizontal to be the 2.1 I worked out.
|\
| \
|  \
----

[kamil9876]

I see. The problem is that once Jay is no longer on the slide, the only force acting on him is gravity, meaning that his motion will be projectile motion and not motion in a straight line. This means you cannot use pythagoras' theorem since his path will now be a parabola.

[TonyHem]

I see. The problem is that once Jay is no longer on the slide, the only force acting on him is gravity, meaning that his motion will be projectile motion and not motion in a straight line. This means you cannot use pythagoras' theorem since his path will now be a parabola.

:crazy2: I get it now!
Thanks again