STAV 2011 + 10

[WhoTookMyUsername]

How did you guys go in these?
What did you find interesting?
Thought some of the questions were quite good... do you guys think it's much harder than VCAA?

[meemz]

stav 2011 lost 6 marks.. found it quite easy, though in structures and materials i dropped 4 careless marks for reading the axes on the graph wrong, lolol. And stav 2010 lost 3 marks it was straight forward... neap 2011 was the hardest i've done by far, losing 8 marks..

[Hutchoo]

Haven't done STAV 2011/2010, but I have done STAV 2012. It was a pretty good exam. I like STAV exams Pretty good imo.

[meemz]

yeah stav 2012 was ideal aswell.. maybe a litlle easier than vcaa, lost 3 marks for it.. did anyone do vcaa 2010? lost 5 marks on it ..

[max payne]

yeah stav 2012 was ideal aswell.. maybe a litlle easier than vcaa, lost 3 marks for it.. did anyone do vcaa 2010? lost 5 marks on it ..

yeah vcaa 2010 seemed trickier than previous years. had a few more 'concept' type questions.

[StumbleBum]

Dropped 5 marks on the 2010 STAV one, was quite easy just made a couple of really stupid working errors.
Also did you find that the solutions were wrong on question 6?

[meemz]

im assuming q6 of motion, and yes it was a mistake. They calculated the area under the graph wrong. Only the area under the decreasing gradient was essential to find change in momentum 

[StumbleBum]

im assuming q6 of motion, and yes it was a mistake. They calculated the area under the graph wrong. Only the area under the decreasing gradient was essential to find change in momentum 

what do you mean only the decreasing part, how would you work it out doing it this way?

I just found the area under the entire graph as the Impulse (60000N.s) and then the impulse is equal to the change in momentum
60000=2200*v
v=27.2m/s

[meemz]

It's asking for the speed before he applied the brakes... Therefore, you exclude before 8 seconds, as he pressed the brakes at 8 seconds, and find the area under the triangle(after 8 seconds) to find change in momentum.. once change in momentum is found, you sub it into the formula : change in p (momentum) = m (v-u).. we know that v is 0 as it comes to rest, if you read the question.. we have the mass, and solve for u..

[StumbleBum]

Ahh i see what you were saying, uhh pretty sure that is wrong though because he applies the brakes from time=0 and at time=8 its just him applying the brakes at a lesser rate until the car comes to rest. So to get the correct answer you have to do the area under the graph from time=0 to time=12 and then that is equal to the change in momentum.

[meemz]

stav 2010 right? question doesn't state that he applies the brakes at t=0.. plus the force is constant for the first 8 seconds ( from the graph) meaning he can't be braking in that time period

[StumbleBum]

Although the graph is labelled "Braking force vs. Stopping time" so the force in the graph is not the net force on the car, but solely the braking force and therefore it is braking from time=0 to time=12 regardless of whether its constant or braking at a decreasing rate the entire area must still be taken.

[meemz]

your method, is the best method.. so you go for it and do it your way.. i asked my teacher today, and he got the same answer as me.. lol

[StumbleBum]

No need to get cynical...
You stated that he was not braking in the first 8 seconds, but if the graph is of the braking force on the car vs time, then how can the car not be braking during that time if it has a constant braking force of 6000N applied to it?

[meemz]

cynical ? hahah, lmao rather realistic if you ask me... i don't know if you did neap 2012, but the first motion question had a friction x distance graph.. the areas where the graph is constant meant no friction just driving force.. i used the same logic for this question..