Recent content by Daedalus13

Thanks for bringing this up, because it is a common mistake that arises when students don't respect the differences between scalars and vectors. Firstly, remember that intensity is not a vector quantity, so I_x and I_y do not make sense. The problem is that the electric field due to the x and...

Thanks to everyone who sat the inaugural BoS Physics Trial! I hope that everyone enjoyed these questions (or at least found them helpful) in preparation for the coming HSC exam. Below are some thoughts I had during the marking of Section 2 Booklet 1. General Use words! You can’t explain...

The constant solutions are asymptotes, so your solution cannot be correct.

I did the approach that 011235 mentions, but got the same answer as you the first time, 358. 402 seems much too large to be plausible if it's 95% chance someone will turn up.

Prove u - p is perpendicular to v with dot product. Let q be any point on v i.e. q = lambda v. Then by pythagoreas |u-p|^2 + |p-q|^2 = |u-q|^2. So |u-q|^2 >= |u-p|^2 and sub in with lambdas

I believe 6 C is incorrect, it should be B (see above).

Also the diagram was 'DRAWN TO SCALE', so 3.2 is much too large.

If you consider f(x)= x^3 and its inverse, the intersection at the origin has the tangents to the graph be perpendicular. So D is also incorrect? Edit: I realised that a vertical tangent means it f'(x) is not defined for the inverse for all real x.

Did anyone get the actual answer for 14c?

I belive it's B because the constant solutions are asymptotes of a particular solution. So -pi/2 and 3pi/2.

On the way down, the velocity is negative (as you define upwards as positive). In this case, an 'resistive force' of 0.1v is still negative i.e. in the same direction as the velocity. So your resistive force is pushing the mass down when it should be pushing it up. You should have a = -g - 0.1v...

I graphed the x function and the axis of symmetry is around 3.3 (using v_0 = 39.1), not quite 3.5. So the solution is valid, with the time to fall longer than the time to go up.

This is what I thought in the exam (to my demise as I made some algebraic errors), but I think the change in sign of the velocity accounts for this change in direction. You end up with the same answer nonetheless (if you don't screw up like I did).

[/QUOTE Is this something you would be penalised for in the HSC? If so, what is the best way to avoid this?

Thanks so much! I never would have thought of that first step.